Kinetic Theory Of Gases

No, the kinetic energy of the molecule does not increase.

Explanation

As we know that kinetic energy of ideal gas is given by

……(i)

Where K. E=kinetic energy of gas

k_B = Boltzmann constant

T = temperature of gas

The Boltzmann constant is a physical constant that relates the average

kinetic energy of particles in a gas with the temperature of the gas.

Since k_B is a constant, kinetic energy of gas is proportional to

temperature of gas. Also, we know

…….. (ii)

Where v= velocity of gas

From equation (i) and equation (ii), we can say that the Kinetic energy of a gas is dependent on two factors-

1. Temperature (K.E. ∝ T)

2. Velocity (K.E. ∝ velocity²)

We can conclude that kinetic energy of gas will change only when there is a change in velocity of gas which in turn will change the temperature of gas. So, if van moves with uniform velocity (i.e. constant speed) no change in kinetic energy or temperature will be observed. Kinetic energy and temperature of the gas will only change when the van either decelerate or accelerate (rate of change of velocity).

1. Cooking gas inside cylinder has approximately 85% of liquid and rest vapors of that liquid.

2. This liquid and vapor is in equilibrium with each other. The pressure in such a system is dependent only on temperature.

3. As the gas is released from the cylinder, the pressure and temperature inside cylinder decreases. To compensate this fall in pressure, phase transition from liquid to gas takes place and some of liquid changes into vapors.

4. For this to happen, system takes heat from surrounding and hence temperature of the system remains constant. And temperature constant means pressure also becomes constant.

5. If phase transition takes place, pressure inside the cylinder will remain same.

6. Hence because of continuous phase transition inside the cylinder, pressure of cooking gas cylinder does not fall appreciably.

No, we cannot except gas inside cooking cylinder to obey ideal gas equation.

Explanation

1. Cooking gas is in liquid form inside cylinder. This means it is under high pressure and low temperature.

2. Ideal gas equation is valid only for gases at low pressure and high temperature. At low pressure and high temperature, the molecules of gas are far apart, and molecular interactions are negligible. Without interaction the gas behaves like ideal gas. So, cooking gas is not an ideal gas as it in liquified form.

Temperature is defined as the average kinetic energy of molecules.

A) No, we cannot define temperature of vacuum.

Explanation

Vacuum is defined as space which contains no matter or space where pressure is so low that no interaction can takes place between any entities or matter. When there are no interactions, no molecule can form as formation of molecules requires Vander Waal force of attraction. So, temperature of a vacuum cannot be defined.

B) No, temperature of single molecules cannot be defined.

Explanation

Temperature for single molecule cannot be defined as we define temperature as the average kinetic energy of all molecules of a gas. And a gas cannot just comprise of single molecule. If that were the case, then there should be no need to do average. So, the answer is no.

1. All the molecules in sample of gas moves with different velocities. Therefore, we average out their velocities to go by our calculation of temperature.

2. The sole reason for averaging is to find that one value of velocity that we can assign to all the molecules and find the temperature of gas as per the definition which is – the average kinetic energy of molecules in a gas is directly proportional to temperature of the gas.

Where K. E=kinetic energy of gas

k_B = Boltzmann constant

T = temperature of gas

3. So yes, temperature of all molecules in a sample of a gas is the same.

Yes, neutron gas is expected to behave much better ideal gas than hydrogen gas.

Explanation

A gas can be called an ideal gas if it follows certain properties.

1) The volume of particles of ideal gas should negligible.

2) There should be no interaction between particles of ideal gas.

Now according to above properties, a neutron gas is expected to behave more like an ideal gas than hydrogen gas because of following reasons:

(i) Neutron is a neutral particle (no charge). So, there will be no interaction between neutron particles. Whereas hydrogen have electron and proton due to which there is possibility of interaction among hydrogen molecules.

(ii) Secondly neutrons are very small in size as compared to hydrogen molecule. So, neutron gas also fulfills the second condition of ideal gas more precisely than hydrogen gas.

No, pressure will not increase.

Explanation

1. Rigid solid is defined as the solids having definite shape and size. A rigid body does not deform under any force or pressure.

2. Pressure of gas inside a container will only change, if the container gets compressed (deform) under the action of force. In our case, the container is rigid cubical container.

3. So, a load of 10kg will not be able to deform the shape and size of container and hence the pressure of the gas will not increase.

No, the molecules of gas will neither collide nor transfer momentum.

First explanation

Kinetic energy of gas is given as

Where K. E=kinetic energy of gas

k_B = Boltzmann constant

T = temperature of gas

So, if T=0K then kinetic energy will be zero. Which also means that molecules of gas will not move at all. Hence, they will neither collide or transfer momentum to walls of container.

Second explanation

One can also understand this from zero pressure point of view.

Pressure of an ideal gas given as

P= nmv² ……... (i)

Where n=number of molecules of gas per unit volume

m=mass of molecule of gas

v= velocity of gas molecule

And momentum is given as

Momentum= massvelocity …...(ii)

From equation (i) and (ii), we can conclude that if pressure is zero,

velocity of molecules will be zero and hence momentum will be zero.

No, we cannot assign temperature for this gas as temperature is not defined for single molecule.

Explanation

The following assumption of kinetic theory will not be valid for single molecule gas:

•A given amount of gas is a collection of large number of molecules that are in random motion colliding with each other and the walls of container.

No, temperature will not decrease.

Explanation

Charles’s law states that for fixed pressure, volume of a gas is proportional to its absolute temperature.

VT …… (pressure=constant)

So, Charles’s law is not even applicable because the pressure of the gas is being reduced in question.

1. Boiling point of a substance increases with increase in pressure. Inside the pressure cooker, pressure is more than atmospheric pressure.

2. We know that boiling point of water at atmospheric pressure is 100.

3. So, boiling point of water is also above 100 inside pressure cooker. This means that food will now cook at higher temperature than 100.

4. So, by increasing the pressure and the boiling point we are reducing the time taken for food to cook. Hence, cooking is faster in pressure cooker than in open vessel.

One should expect more evaporation.

Explanation

1. Evaporation is the transition from liquid state to gaseous state. To do so heat is provided so that forces holding molecule together in liquid state are weakened and vapor state can be achieved.

2. If molecules are not allowed collide among themselves this means that interaction between them is very weak. Hence, more evaporation will take place.

Yes, it is possible to boil water at 30

Explanation

1. Boiling point decrease with decreases in pressure.

2. So, if the pressure of the flask containing water is reduced up to value so that boiling point is reduced from 100 to 30 then, the water will start boiling.

3. But since now the temperature at which the water has started boiling is very low, flask will not be hot.

After we come out of river water sticks to our body. That water evaporates by taking heat from our body. Thus, there is a transfer of heat from our body to water droplets. Therefore, temperature of our body reduces, and we feel cold.

Kinetic energy of gas is dependent on temperature of gas. Mathematically it is given as

Where K. E=kinetic energy of gas

k_B = Boltzmann constant

T = temperature of gas

From above formula kinetic energy is directly proportional to temperature. Since k_B is constant therefore, kinetic energy is constant and same for all gases at a given temperature.

1. According to kinetic theory of ideal gas, molecules of ideal gas should be in incessant random motion and constantly colliding with each other and with the wall of the container in which they are kept.

2. This can only happen when they have large velocities or kinetic energy. And we know that temperature and kinetic energy are directly proportional to each other.

Where K. E=kinetic energy of gas

k_B = Boltzmann constant

T = temperature of gas

So, large kinetic energy means high temperature.

3. Another postulate of kinetic theory states that size of molecule should be very small as compared to the volume of gas.

4. This can be achieved at low pressure because at low pressure concentration of gas is very low (less number of molecule in large volume or space). So, gas behaves as ideal gas at low pressure and at high temperature.

1. Molecules of ideal gas have negligible interaction between them. This means that force acting due to each other while they are in motion is almost zero.

2. In such cases where force acting on particle is zero, particle moves in uniform motion and in straight line according to newton’s first law of motion.

3. So, in kinetic theory of ideal gas, molecules of gas are moving in straight line. Hence, they will only have translational kinetic energy.

As we know that kinetic energy of ideal gas is given by

Where K. E=kinetic energy of gas

k_B = Boltzmann constant

T = temperature of gas

Since k_B is a constant therefore, kinetic energy of gas is proportional to temperature of gas.

Rms speed at given temperature is given as

v_rms =

where R=gas constant whose value is 8.31 J/mol K.

T=temperature of gas.

M= molar mass of molecule of gas

Therefore, at a given temperature rms speed of gas is inversely proportional to square root of molar mass of molecule of that gas.

Out of the four-option molar mass of hydrogen molecule is least

i.e. 2 amu. So rms speed of hydrogen will be maximum.

From the graph, we can see that slope of straight line with temperature

T₁ is greater than slope of straight line with temperature T_2.

Slope of straight line is given as:

Slope =

In our graph, slope is

where P=pressure

= density of gas

So, ratio for T₁ is more than T_2.

Rms speed of gas is given as

where R=gas constant whose value is 8.31 J/mol K.

T=temperature of gas.

M= molar mass of molecule of gas

We also know that ideal gas equation is

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

N=number of moles of gas

So, we can write .

Putting the value of RT in equation (i),we get

nM= total mass of gas ‘m’

and density =

Putting the value of density in equation (ii)_, we get

Since ratio for T₁ is more than T₂ therefore rms speed for ideal gas at temperature T₁ is more than temperature T₂. And if rms speed is more for temperature T₁ then T₁ > T₂ from formula

v_rms =

Root mean square velocity of gas is given as

v_rms =

where R=gas constant whose value is 8.31 J/mol K.

T=temperature of gas.

M= molar mass of molecule of gas

Squaring both side of equation (I) we are removing the square root and

we will get mean square velocity which is,

So, mean square speed is directly proportional to temperature as for a

given gas its molar mass will not change.

For a single molecule speed will be same whether it is rms speed or average speed.

Root mean square velocity of gas is given as

v_rms =

where R=gas constant whose value is 8.31 J/mol K

T=temperature of gas

M= molar mass of molecule of gas

rms speed of oxygen = …(i)

rms speed of hydrogen= …(ii)

Given:

v_rms(o)=500m/s

and we know molar mass of oxygen =M(o)=32

molar mass of hydrogen=M(H)=2

Diving equation (i) and (ii) we get

Since speed of both gases must be calculated at same temperature this equation will reduce to

m/s ()

Root mean square velocity of hydrogen is 2000m/s.

Given:

Pressure P=200kPa

1kPa= Pa

We know that ideal gas equation is

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

So,

=200Pa ……(i)

Now according to question half of the gas is removed. That means number of moles left behind are also halved.

Therefore, number of moles left behind n’=

So new pressure will be P’=

Putting the value of n’ in above equation we get

From equation (i) we can write P’ as

Pa=100kPa.

New pressure will be 100kPa.

Root mean square velocity of gas is given as

v_rms

where R=gas constant whose value is 8.31 J/mol K.

T=temperature of gas.

M= molar mass of molecule of gas

Given,

Root mean square of oxygen

v_rms

Molar mass of oxygen molecule M=32. So,

When temperature is doubled, oxygen molecule dissociates into oxygen atom.

Molar mass of oxygen atom M’=16

New temperature T’=2T

Then rms speed of oxygen atom becomes

Multiplying and diving above equation by 2 we get,

New rms speed of oxygen will ve 2v.

We know that ideal gas equation is

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas.

Gas constant R=kN_A

Where k=Boltzmann constant

N_A=Avogadro number

So, we can ideal gas equation as

PV=nkN_AT

Now, we know that

n=number of molecules of gas ……(ii)

Therefore, from equation (i) and (ii)

Number of molecules of gas=

From the graph, we can conclude that pressure is directly proportional to temperature because it is straight line graph between pressure and temperature.

We know that ideal gas equation is

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas.

From above equation, we can see that pressure will be directly proportional to temperature only when volume V is also a constant.

So, only for isochoric process (a process where volume is kept constant) pressure will be directly proportional to temperature.

1. Saturated air means air having maximum humidity or moisture. If we keep on adding moisture to saturated air the extra moisture will condense. And saturated air will become unsaturated air at that temperature.

2. Since amount of liquid is continuously decreasing that means evaporation is taking place.

3. Now during vaporization, moisture is being continuously added to air above liquid. Temperature is changing continuously.

4. So, vapors in remaining part is unsaturated as air contains extra moisture.

1. Since the amount of liquid remains constant that means no vapors or moisture is being added to air or vapors above it.

2. Saturated air means air having maximum humidity or moisture. If we keep on adding moisture to saturated air the extra moisture will condense. And saturated air will become unsaturated air at temperature.

3. Since no moisture is being added, remaining vapors must be saturated.

1. Maximum pressure attainable by any liquid is saturation vapor pressure. Saturation vapor pressure is the pressure exerted by vapor above the surface of liquid when the air has become saturated.

2. Initially when vapor is being injected at a uniform rate the pressure inside the chamber will increase.

3. But after a certain limit air inside vessel will become saturated and then if we add more vapor to air it will condense.

4. At that point pressure will stop increasing and will become constant as pressure has reached the value of saturated vapor pressure.

So, pressure will increase first and then will become constant.

1. Maximum pressure attainable by any liquid is saturation vapor pressure. Saturation vapor pressure is the pressure exerted by vapor above the surface of liquid when the air has become saturated.

2. When the air become saturated, pressure of the gas depends completely on temperature and not on volume of liquid.

3. So, in our case the water is same in both the vessel i.e. volume of water is same, and temperature of water is also constant.

4. Hence, pressure in both the vessel will be same.

1. According to kinetic theory of ideal gases, molecules of a gas are in incessant random motion, colliding against each other and with the walls of the container. All these collisions are perfectly elastic collision.

2. In perfectly elastic collision, total kinetic energy of gas is conserved.

3. Here, we have a mixture of oxygen and hydrogen gas. Therefore

K.E of oxygen + K.E of hydrogen=constant

4. So, if kinetic energy of oxygen molecules increases, then kinetic energy of hydrogen molecules decreases or vice-versa so that, sum of both kinetic energies remains constant.

5. Therefore, both option (c) and (d) are correct.

In kinetic theory of ideal gas, the average energy is given by

Where R=gas constant=8.31Jmol^-1K^-1

T=temperature of gas

M=molar mass of gas

1. From the formula for average speed, it can be seen that .

2. Molar mass of hydrogen molecule is 2 amu and molar mass of oxygen molecule is 16 amu.

3. So, molar mass of oxygen molecule is greater than molar mass of hydrogen molecule.

4. Therefore, average speed of oxygen molecule will be less than average speed of hydrogen molecule.

1. We know that momentum is a vector quantity defined as product of mass and velocity of particle in motion.

2. According to kinetic theory of ideal gas, molecules of gas are in random motion.

3. Due to this random motion, velocity on an average will be zero. This is because velocity is a vector quantity, having both direction and magnitude.

4. Because velocity have direction, in random motion all the components of velocity will cancel out each other and it will be zero on an average.

5. Thus, average momentum will also be zero.

We know ideal gas equation is

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

So, we can write

According to question, n, V and T are constant.

R= gas constant is universal constant.

So, from equation (i) Pressure will be same for all ideal gas.

1. Average momentum of a molecule in sample in ideal gas is zero.

2. We know that momentum is a vector quantity defined as product of mass and velocity of particle in motion.

3. According to kinetic theory of ideal gas, molecules of gas are in random motion.

4. Due to this random motion, velocity on an average will be zero. This is because velocity is a vector quantity, having both direction and magnitude.

5. Because velocity have direction, in random motion all the components of velocity will cancel out each other and it will be zero on an average.

6. Thus, average momentum will also be zero.

7. Therefore, average momentum is independent of all the quantities mentioned options.

1. Avogadro's law states that, "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."

2. So, from Avogadro law, at same temperature all the ideal gases will have same volume.

3. As we know that kinetic energy of ideal gas is given by

……(i)

Where K. E=kinetic energy of gas

k_B = Boltzmann constant

T = temperature of gas

4. Temperature is a quantity that is dependent on number of molecule in kinetic theory of gases.

5. Again, from Avogadro law, at same temperature all the gases will have same number of molecules which is equal to Avogadro number.

6. In 1 mole of any gas, number of molecules are 6.02310²³.

7. Therefore, kinetic energy of 1 mole, will be same for all ideal gas at same temperature as all will contain same number of molecules.

In quantity ,

M=mass of the gas

k= Boltzmann constant

t= temperature

p=pressure

V=volume

We know ideal gas equation

pV=nRt

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

t=temperature

n=number of moles of gas

p=pressure of gas.

Gas constant R=kN_A

Where k=Boltzmann constant

N_A=Avogadro number

So, we can ideal gas equation as

pV=nkN_At

Now, we know that

n=number of molecules of gas ……(ii)

Therefore, from equation (i) and (ii)

Number of molecules of gas= …. (iii)

Using equation (iii) quantity becomes

Hence, quantity depends on the nature of gas.

STP means standard temperature-273.15K and pressure101.325 kPa.

Given

Pressure P=1.01Pa

Number of moles n=1

Temperature T=273.15K

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

So, we can write

The volume of 1 mole of an ideal gas at STP=0.0224m³.

STP means standard temperature-273.15K and pressure

101.325 kPa.

Volume of ideal gas at STP= 22.4L

Number of molecule in 22.4L of ideal gas at STP=Avogadro number =6.022

Now we know that

1Litre=10³cm³

22.4L=22.410³cm³

Number of molecule in 22.410³cm³ of ideal gas at STP=6.022

Therefore,

Number of molecules in 1cm³ of ideal gas at STP==.

Given

Volume of ideal gas V=1cm³

1cm=m

V=m³

Temperature of ideal gas=0

T(K)=T ()+273.15

T=T(K)=0+273.15=273.15K

Pressure of ideal gas P=mm of Hg

1mm of Hg= 133.32Pa

P=mm of Hg=133.32Pa

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant=8.31J/molK

T=temperature

n=number of moles of gas

P=pressure of gas.

So,

Number of molecules = Avogadro numbernumber of moles

Number of molecules=n

=

=3.538

Therefore,

Number of molecules in 1 cm³ of an ideal gas at 0°C and at a pressure of 10^–5 mm of mercury is 3.538.

STP means standard temperature-273.15K and pressure

101.325 kPa.

Given

Volume of oxygen gas=1 cm³

We know that

Volume of oxygen gas at STP= 22.4L

1Litre=10³cm³

22.4L=22.410³cm³

Therefore, volume of oxygen gas =22.410³cm³

We know that 22.4L of O₂ contains 1 mol O₂ at STP.

22.410³cm³ of O₂ =1 mol O₂

Therefore,

1cm³ of O₂ = mol of O₂

I mol of O₂ =32 grams of O₂

mol of O₂ = grams of O₂=1.43grams

1g=mg

The mass of 1 cm³ of oxygen kept at STP=1.43mg.

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas.

Given

Masses of both the gas is equal. Therefore, number of moles of both the gas is equal. So, we can write

n₁ =n₂ =n

Volume of first gas V₁=V_o

Volume of second gas V₂=2V_o

Temperature of first gas T₁=300K

Temperature of second gas T₂=600K

Let pressure of first gas =P₁

Pressure of second gas=P₂

Applying ideal gas equation for both the gases

P₁V₁=n₁RT₁

…(I)

P₂V₂=n₂RT₂

…. (II)

Since n₁ =n₂ =n

Therefore

Rearranging the above equation

P₁:P₂ =1:1

So, the ratio of pressure gas in two vessels is 1:1.

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas.

Given

Volume of gas=250cc

1cc=1cm³m³

V=250m³

Pressure P=10^–3 mm of mercury

1mm of Hg= 133.32Pa

P= mm of Hg=133.32 Pa

Temperature T=27

T(K)=T ()+273.15

T=T(K)=27+273.15=300.15K

From ideal gas equation, we can write

Number of molecules = Avogadro numbernumber of moles

Number of molecules=n

=

Number of molecules in electric bulb=.

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas.

Given

Maximum pressure P_max=P₂=1.0 × 10⁶ Pa

Pressure of gas P₁=8.0 × 10⁵ Pa

Temperature of gas T₁=300K

Since the volume has not been changed therefore,

V₁=V₂=V

Hence number of moles will also be same n₁=n₂=n

Temperature at which the cylinder will break=T₂

Since n₁ =n₂ =n

Therefore,

T₂ =375K

The temperature at which the cylinder will break=375K.

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas.

Given

Volume V=0.02 m³

Temperature T=300K

Mass of hydrogen gas = 2g

Number of moles n =

Molar mass of hydrogen=2amu

mol

From ideal gas equation, we can write

Pa

Pressure in the vessel is Pa.

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas

STP means standard temperature-273.15K and pressure 101.325 kPa.

Given:

T=273.15K

P=101.325Pa

Density of ideal gas 1.25 × 10^–3 g cm^–3

1gcm^-3=10³kgm³

kgm³

Number of moles n ==

Density

From ideal gas equation, we can write

Molecular weight of gas is 2.3810^-2gmol^-1

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas

Number of moles n ==

Density

Given:

Temperature of Shimla T₁=15.0°C

T(K)=T ()+273.15

T₁=T(K)=15+273.15=288.15K

Pressure of Shimla P₁ = 72.0 cm of mercury

Temperature of Kalka T₂ = 35.0°C

T(K)=T ()+273.15

T₂=T(K)=35+273.15=308.15K

Pressure of Kalka P₂=76.0 cm of mercury

Substituting the value of n and in ideal gas equation, we get

=

So,

Taking the ratio of equations (i) and (ii),

The ratio of air density at Kalka to the air density at Shimla is 0.98.

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas

Given

Volume of first part=V

Volume of second part=3V

Initially separator had divided cylinder in two equal parts so, number of moles in both the parts will be same.

n₁=n₂=n

Since the walls of separator is diathermic, the temperature of both the parts will always be same.

T₁=T₂=T

Pressure of part 1

P₁=

Pressure of part 2

P₂==

Diving P₁ and P₂, we get

P₁:P₂ =3:1

The ratio of the pressures in the two parts of the vessel is 3:1.

We know that rms speed of gas is given by

Where R=gas constant 8.31J/molK

T=temperature of gas

M=molar mass of gas

Given

Temperature T=300K

Molar mass of hydrogen=2g/mol

Therefore,

Now in second part of question speed is doubled i.e. 21932.6ms^-1

Let temperature at this speed be T₁

So, using the same formula of rms speed

Squaring both sides of above equation

Temperature of the gas when speed is doubled is 1200K which is 4 times the pervious temperature.

STP means standard temperature-273.15K and pressure 101.325 kPa.

We know that rms speed of gas is given by

Where R=gas constant 8.31J/molK

T=temperature of gas

M=molar mass of gas

Given

Temperature T=273.15K

Pressure P=101.325Pa

Mass =0.177g =0.17710^-3kg

Volume = 1000cm³

1cm=10^-2m

1cm³=10^-6m³

1000cm³=10^-3m³

Density =

We know that ideal gas equation is

PV=nRT

Where V= volume of gas

R=gas constant=8.31J/molK

T=temperature

N=number of moles of gas

So, we can write .

Putting the value of RT in v_rms we get

v_rms =

nM= total mass of gas ‘m’

and density =

Putting the value of density and m in v_rms, we get

v_rms =

Putting the value of P and in above equation we get

The rms speed of the gas molecules at STP is 1310.4ms^-1.

We know that kinetic energy of ideal gas is given by

…. (I)

Where K. E=kinetic energy of gas

k_B = Boltzmann constant=1.38 × 10^–23 J K^–1.

T = temperature of gas

In kinetic theory of ideal gas, molecule of gas is moving in straight line.

Hence, they will only have translational kinetic energy.

Given

K. E=0.040 eV

We know that

1 eV = 1.6 × 10^–19J

0.040eV=0.040J

From equation (I) we can write

Therefore, the temperature of gas is 309.2K.

In kinetic theory of ideal gas, the average energy is given by

Where R=gas constant=8.31Jmol^-1K^-1

T=temperature of gas

M=molar mass of gas

Given

Temperature T=300K

Molar mass of oxygen=32amu=32g/mol=3210^-3 kg/mol

Therefore

We know that

Distance here is given as diameter of earth

Now radius of earth = 6400km=6400000m

Diameter=2radius

So, diameter of earth =26400000m

So, time taken

1 hour=6060 seconds=3600seconds

So, 28747.83s =h = 7.98h8h

So, average time taken by oxygen molecule to travel a distance

equal to the diameter of the earth is 7.98h8h.

In kinetic theory of ideal gas, the average energy is given by

Where R=gas constant=8.31Jmol^-1K^-1

T-temperature of gas

M=molar mass of gas

Given

Temperature T=0

T(K)=T ()+273.15

T=T(K)=0+273.15=273.15K

Mass of helium molecule m =6.64 × 10^–27 kg

We know that,

Gas constant R=k_BN_A

Where k_B= Boltzmann constant = 1.38 × 10^–23 J K^–1.

And N_A=Avogadro number=6.02310^-23 mol^-1

Molar mass of gas molecule M= Avogadro numbermass of gas molecule

M=N_Am

So average velocity becomes

We know that

Momentum=massvelocity

Momentum=6.64 × 10^–271202.31=8kgms^-1

Average magnitude of momentum of helium at 0 is 8kgms^-1.

In kinetic theory of ideal gas mean speed also known as average

speed is given as

Where R=gas constant=8.31Jmol^-1K^-1

T=temperature of gas

M=molar mass of gas

Given

Let temperature of hydrogen gas =T(H)

Temperature of helium gas= T(He)

Molar mass of hydrogen gas =2amu

Molar mass of helium gas =4amu

Squaring both sides

the ratio of the temperature of the hydrogen sample to the temperature of the helium sample is 1:2.

In kinetic theory of ideal gas mean speed also known as average speed is given as

Where R=gas constant=8.31Jmol^-1K^-1

T-temperature of gas

M=molar mass of gas

Molar mass of hydrogen=2amu=2g/mol=2kg/mol

Escape speed of earth is the speed given to projectile so that it escapes the gravitational field of earth. It is given by formula

Where G=universal gravitational constant

M_E=mass of earth

R=radius of earth

We also know that acceleration due to gravity g is

Multiplying and dividing equation (II) by R

Putting the value of g from equation (III) to above equation we get

According to question v_mean is equal to v_e

So, from equation (I) and (IV) we get

Radius of earth R=6400km=6400000m

g=9.8m/s

Squaring both sides

Temperature at which the mean sped of the molecules of hydrogen gas equals the escape speed from the earth is 11800K.

In kinetic theory of ideal gas, mean speed also known as average

speed is given as

Where R=gas constant=8.31Jmol^-1K^-1

T=temperature of gas

M=molar mass of gas

Molar mass of hydrogen molecule M(H)=2 amu

Molar mass of nitrogen molecule M(N)=28 amu

Mean speed of hydrogen molecule=

Mean speed of nitrogen molecule=

Temperature of both the gases is same.

Dividing equation (II) and (III) we get

The ratio of the mean speed of hydrogen molecules to the mean speed of nitrogen molecules in a sample containing a mixture of the two gases is 3.74.

In kinetic theory of ideal gas, the average energy is given by

Where R=gas constant=8.31Jmol^-1K^-1

T=temperature of gas

M=molar mass of gas

We know that,

Gas constant R=k_BN_A

Where k_B= Boltzmann constant = 1.38 × 10^–23 J K^–1.

N_A=Avogadro number=6.02310^-23 mol^-1

Molar mass of gas molecule M= Avogadro numbermass of gas molecule

M=N_Am

So average velocity becomes

Rms speed of gas molecule is given by

Where R=gas constant 8.31J/molK

T=temperature of gas

M=molar mass of gas

Putting the value gas constant R=k_BN_A

So rms speed becomes

M=N_Am

Therefore,

Let the mass of molecule in left part=m₁

Mass of molecule in right part=m₂

According to question, the rms speed of the molecules in the left part equals the mean speed of the molecules in the right part.

So, from equation (I) and(II) we get

Since the walls of separator is diathermic therefore temperature of both the parts will be same.

Squaring the above equation, we get

The ratio of the mass of a molecule in the left part to the mass of a molecule in the right part is 1.17.

Number of collision per second means frequency of collision.

In kinetic theory of ideal gas, the average energy is given by

Where R=gas constant=8.31Jmol^-1K^-1

T-temperature of gas

M=molar mass of gas

Molar mass of hydrogen =2 amu=2kg/mol

Given

Distance between successive collision 1.38 × 10^–5 cm

1.3810^-8m

Time between two collisions

Frequency of collision =

Number of collision per second means frequency of collision which is equal to 1.2310¹¹.

(a)

In kinetic theory of ideal gas, mean speed also known as average speed is given as

Where R=gas constant=8.31Jmol^-1K^-1

T=temperature of gas

M=molar mass of gas

Given

T=300k

Molar mass of hydrogen gas =2amu=2g/mol=2kg/mol

(b)

Let velocity be u= from part (a)

From figure we can see that

Total Momentum in vertical direction

mucos45-mucos45=0

Total momentum in horizontal direction

musin45-(-musin45)=2musin45=2mumu

Total change momentum of 1 molecules =mu

Total change momentum of n molecules =nmu

We know that,

Let ‘t’ be the time taken to changing the momentum.

So, force per unit second due to 1 molecule=mu

Force per unit second due to n molecule=nmu

Given

Pressure by n molecule=10⁵ Pa

Area=1m²

Pressure =

We know that

Mass of 6.023 of hydrogen molecule=2kg

Mass of 1 hydrogen molecule m =kg

Therefore

Number of molecules strike each square meter of the wall per second=1.2

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas

Given

Pressure at temperature 20 P₁=200Pa

Volume at temperature 20 =V₁

Increase in volume =2%V₁

Volume at temperature 40 V₂=V₁+2%V₁

V₂=V₁+0.02V₁=1.02V₁

20=293.15K

40=313.15K

From ideal gas equation, we can write

Since, tube got expanded by the end of the day, only volume will change. Number of moles remains the same, before and after expansion as no new gas has been added. So, the product of nR will be same before and after expansion.

Pressure of the tube at 40⁰C is 209.45kPa.

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas

Given

Volume inside jar V₁= 1.0 × 10^–3 m³

Pressure inside jar P₁ =1.5 × 10⁵ Pa

Temperature inside jar T₁=400K

Pressure of surrounding P₂ =1atm=1.0Pa

Temperature of surrounding T₂=300K

Let volume of oxygen at T₂ and P₂ =V₂

When jar is in equilibrium with surrounding, temperature and pressure of oxygen gas inside jar will T₂ and P₂.

Number of moles will be same inside the jar, before and after equilibrium as no new oxygen gas has been added. Just temperature and pressure has been changed. Due to which volume will change.

Assuming there is no leak in jar, applying ideal gas equation before and after equilibrium, we get

Now if we consider leak,

Volume of gas leaked =V₂-V₁

= (1.125-1)m³

= 1.25m³

If n₂ are number of moles leaked out, then

Mass of the gas leaked out =n₂molar mass of oxygen molecule

Molar mass of oxygen molecule=32g/mol

Mass of gas leaked out=0.00532=0.16g

The mass of the gas that leaked out =0.16g.

Given

Radius of bubble at the bottom of deep river R₁=2.0mm=2.0m

Depth of the river h=3.3m

Density of water = 1000 kg m^–3

We know that

Pressure at depth inside a fluid is related to atmospheric pressure by relation

P₁=P_a + hg

Where P₁ =pressure at depth h

P_a=atmospheric pressure=1.0 × 10⁵ Pa

g=acceleration due to gravity=9.8ms^-2

=density of fluid.

So,

P₁=1.010⁵+3.310009.8=1.3210⁵Pa

Temperature is same for both water at the bottom and the water at the surface. So, we can apply Boyle’s law which says that PV=constant, when temperature is constant.

Let V₁ be the volume of air bubble at bottom of deep river

V_a be the volume of air bubble at surface of river

Where R_a = radius of bubble at surface of river

So, according to Boyle’s law

P₁ V₁=P_a V_a

R_a=2.210^-3m

Radius of the air bubble at the surface of river is 2.210^-3m.

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant=8.31Jmol^-1K^-1

T=temperature

n=number of moles of gas

P=pressure of gas

Given

Pressure inside the tyre P₁=2atm=2Pa

Volume at P₁, V₁=0.002m³

Reduced volume V₂=0.0005m³

Temperature remains constant so T₁=T₂=300K

Let when the gas is leaked out the pressure P₂ becomes equal to atmospheric pressure. So P₂=1.0Pa.

Number of moles initially n₁

Similarly

Final number of moles n₂

So, number of moles leaked out will be n₁-n₂=0.16-0.02=0.14.

Given

Mass of helium =0.040g

Molar mass of helium=4g/mol

Number of moles n=

Number of moles for helium n=

Temperature T₁=100

T(K)=T ()+273.15

T=T(K)=100+273.15=373.15K

Internal energy U in kinetic theory is given as

Where C_v= molar specific heat capacity

N= number of moles

T=temperature of gas

Also, internal energy depends only the temperature of the gas.

Helium is a monoatomic gas and for monoatomic gas

So, C_v for helium is Jmol^-1K^-1

Increase in internal energy is given in question as 12J. That means

U₂-U₁= nC_v(T₂-T₁)

Since the gas has not been changed, just expanded no change in molar specific heat capacity and number of moles of gas.

Putting the value of change in internal energy and T_1,

12=0.0112.45(T₂-373.15)

The temperature at which the internal energy is increased by 12J is 469.53J =196.38.

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant=8.31Jmol^-1K^-1

T=temperature

n=number of moles of gas

P=pressure of gas

So ………. (I)

Now differentiating the ideal gas equation, we get

PdV + VdP=nRdT …………. (II) (we have applied product rule for

differentiation of PV)

Now as given in question the ideal here follows and additional law which is PV²=constant.

So, differentiating this additional law as well we get

2PVdV + V²dP=0

Taking V as common we get

2PdV + VdP=0 ……… (III)

Subtract equation (III) from (II)

2PdV + VdP - PdV - VdP = -nRdT

PdV=-nRdT

From equation (I), substitute the value of P in above equation we get

Integrating equation (IV) from limits V to 2V and T1 to T2

We know . Applying this formula

Where we have applied the property of ln which is

ln(a)-ln(b)=ln(a/b)

So, the temperature at which the gas expands is half of the initially temperature.

Given

Mass of oxygen gas =1.60g

Mass of nitrogen gas =2.80g

Temperature of vessel=300K

Volume of vessel=0.166m³

We know that

Number of moles n=

Molar mass of oxygen= 32g/mol

Molar mass of nitrogen=28g/mol

Number of moles of oxygen n₁ =

Number of moles of nitrogen n₂=

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant=8.31Jmol^-1K^-1

T=temperature

n=number of moles of gas

P=pressure of gas

In a mixture of non-interacting ideal gases, the pressure that a gas in a mixture of gases would exert if it occupied the same volume as the mixture at the same temperature is called the partial pressure of that gas.

Partial pressure of oxygen gas

Partial pressure of nitrogen gas

According to Dalton’s law of partial pressure, the total pressure of a mixture of ideal gas is the sum of partial pressures.

So, total pressure

P=P_o+P_N =750+1500=2250Pa

The pressure of the mixture of oxygen and nitrogen gas is 2250Pa.

Given

Height of vertical cylinder=100cm=1m

Pressure P₁=75cm of Hg=0.75m of Hg

1mm of Hg = hg Pa

Where h= height of mercury column =1mm=0.001m

= density of mercury

g= acceleration due to gravity

So,

P₁= 0.75m of Hg= 0.75g Pa

Let h be height of mercury above the piston. When mercury is poured over piston the piston will move down and gas inside vessel will get compressed.

So, let the pressure of gas when mercury is poured be P₂

So,

P₂=P₁+hg=0.75g+ hg

Let the circular area of cylinder be A.

Then, volume of gas before mercury was poured V₁=Aheight of cylinder

V₁=A1=A

Height of cylinder when mercury was poured =(1-h) m

Volume of gas after mercury was poured V₂=A(1-h)

Since it is given in question that temperature has not being changed so we can apply Boyle’s law which states that PV=constant, if temperature is constant.

P₁V₁=P₂V₂

0.75gA=0.75g+ hg A(1-h)

Taking gA common from both side of equation we get

0.75=(0.75+h) (1-h)

0.75=0.75+h-0.75h-h²

h²-0.25h=0

h-0.25=0h=0.25m

The maximum height of the mercury column that can be put on the piston is 25cm.

Let the partial pressure of gas A and B be P’_A and P’_B respectively.

Given:

Pressure, temperature and volume of gas A P_A T_A, V

Pressure, temperature and volume of gas B P_B, T_B, V

We know that ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

In a mixture of non-interacting ideal gases, the pressure that a gas in a mixture of gases would exert if it occupied the same volume as the mixture at the same temperature is called the partial pressure of that gas.

Using this definition volume of gas, A when gas B in not present, is 2V and temperature T.

So, from ideal gas equation

P’_A2V=nRT

Also

P_AV=nRT_A

Equating nR from both the above equation

Doing the same above procedure for gas B

P’_B2V=nRT

Also

P_BV=nRT_B

According to Dalton’s law of partial pressure, the total pressure of a mixture of ideal gas is the sum of partial pressures.

Given

Volume of container V₁=50cc=5010^-6m³

Molecular mass of air in container M= 28.8g

Pressure of air P₁= 100kPa=10⁵Pa

(a) We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant=8.31Jmol^-1K^-1

T=temperature

n=number of moles of gas

P=pressure of gas

In first case the air is kept in container having ice. So, temperature in case will be T₁ =0=273.15K

Number of moles n= …. (1)

Number of moles n= …. (2)

Equating (1) and (2) we get

So, mass of air when temperature is 0 is 0.0635g.

(b) Now in second case the container having air is kept in a bath having boiling water. So, temperature will be T₂=100=373.15K.

Since, now temperature is 100 therefore, some of the air will be expelled as air will expand but the volume of container is fixed. So, some of the air will go out of the container as container is open.

So, first we will calculate the mass of air expelled from container and then we will subtract it from the original volume V₁ to get the mass of remaining air.

Pressure will be same as before, as the air is still open to atmosphere. So P₂=P₁.

Let the volume of expanded gas be V₂. Number of moles in volume V₂ be the same as before because no extra gas is added. It has just expanded.

As P₂=P_1, therefore

Volume of gas expelled out the container

V =V₂-V₁=

Number of moles of expelled gas

So, the mass of gas remaining in the container

=m’-m=0.0635-0.017=0.0465g

So, the mass of gas when temperature is 100 is 0.0456g.

(c) Now the container is kept in ice bath i.e. temperature 0 and container is closed. So, now the pressure will change.

Number of moles left =

Applying ideal gas equation

Pressure of gas when lid is closed, and temperature is 0 is 73.1kPa.

Let the curved surface area of tube be A.

Volume =areaheight

Given

Initial length of trapped air=20cm =0.2m

Length of mercury column=10cm=0.1m

So, Mercury column pressure =0.1g Pa

Initial volume of air trapped V₁=0.2A

Atmospheric pressure =75cm of Hg=0.75m of Hg

1mm of Hg = hg Pa

Where h= height of mercury column =1mm=0.001m

= density of mercury

g= acceleration dur to gravity

So, atmospheric pressure= 0.75m of Hg= 0.75g Pa

Let the pressure of the trapped air when closed end of the tube is upward be P₁. Now, pressure of the mercury and trapped air will then be equal to atmospheric pressure.

P₁+0.1g=0.75g

P₁=0.65g

When the tube is inverted such that closed end is downward then pressure of trapped air will be P₂.

P₂=atmospheric pressure + mercury column pressure

P₂=0.75g+0.1g=0.85g

Let length of air column at P₂=x

Volume of air trapped will be V₂=A x

Now temperature in both cases will remain same, as no heat is being either added or abstracted from tube.

Applying boyle’s law,

P₁V₁=P₂V₂

0.65g0.2A=0.85gAx

The length of the air column trapped when the tube is inverted so that the closed-end goes down is 0.15m=15cm.

Let the curved surface area of tube A

Given

Length of mercury column=10cm=0.1m

Length of tube =100cm=1m

Pressure of mercury column P₁=76cm of Hg=0.76m of Hg

Temperature of mercury column T₁=27=300.15K

Temperature of air at cooler side T₂=0=273.15K

Temperature of air at hotter side T’₂=127=400.15K

Let the length of air column at cooler side be

The length of air column at hotter end be

Volume of cooler air =A

Volume of hotter air =A

Volume of mercury column = V

Pressure of cooler air =P₂

Pressure of hotter air =P’₂

We know that ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

Applying ideal gas equation between cooler air and mercury column

Applying ideal gas equation between hotter air and mercury column

Under equilibrium condition the pressure P2 and P’2 will be same

Now length of entire tube

x+y+0.1=1

y=0.9-x

Substituting the value of y in equation (i)

(0.9-x)273.15=400.15x

So, the length of air column on the cooler side is 0.365m=36.5cm.

Let curved surface area of tube =A

Given

Length of air column=43cm=0.43m

Length of mercury column=20cm=0.20m

Pressure due to mercury column==0.2m of Hg

Atmospheric pressure=P_a=0.76m of Hg

Let the pressure of air column before titling =P₁

So P₁=P_a +P_H

P₁=0.76+0.2=0.96m of Hg

Volume =areaheight

Volume of trapped air V₁=Alength of air column=0.43A

If the tube is titled through an angle 60^o only pressure of mercury column will get affected and not the atmospheric pressure.

So, change in P_H will be

P’_H=P_Hcos60^o = 0.20.5 =0.1m of Hg

So now the pressure of air column will become P₂

P₂=P_a+P’_H=0.76+0.1=0.86m of Hg

Then volume will change. Let it now be V₂=lA where l is new length of air column.

It is given in question that the temperature remains same. So, according to boyle’s law which states that PV=constant when temperature is constant, we can write,

P₁V₁=P₂V₂

Length of the air column will become 0.48m=48cm

Let the initial pressure of the chamber A and B be P_a and P_b respectively.

Let the final pressure of the chamber A and B be P’_a and P’_b respectively.

Let the curved surface area of tube be A

Given:

Length of chamber A=20cm=0.2m

Length of chamber B=10cm=0.1m

Volume =areaheight

Initial volume of chamber A=V_a=0.2A

Initial volume of chamber B=V_b=0.1A

Initial Temperature of chamber A=T_a=400K

Initial Temperature of chamber B=T_b=100K

For first (momentary) equilibrium, pressure of both chamber will be same.

P_a=P_b

Let the final temperature at equilibrium be T.

We know that ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

Then,

For chamber A, number of moles, before and after final equilibrium will be same as no new gas has been added. So, applying ideal gas equation, before and after final equilibrium and equation nR, we get,

Similarly, for chamber B

At second equilibrium pressures on both sides will be same again.

P’_a=P’_b

Now P_a=P_b so,

V’_b =2V’_a ……(iii)

Volume of chamber A plus volume of chamber B will be equal to total volume. So,

V’_b+V’_a=V=0.3A

2V’_a+V’_a=0.3A (from (iii))

3V’_a=0.3A

V’_a=0.1A

Now we know that volume=lengtharea. So,

V’_a=lA

Where l=length of chamber A after equilibrium

lA=0.1A

l=0.1m=10cm

length of chamber A after equilibrium is 10 cm.

Let P be the pressure and n be the number of moles of gas inside the vessel at any given time.

As mentioned in question, pressure is decreasing continuously. So, suppose a small amount of gas ‘dn’ moles are pumped out and the decrease in pressure is ‘dP’.

So, pressure of remaining gas =P-dP

Number of moles of remaining gas =n-dn

Given

The volume of gas =V_o

Temperature of gas=T

We know that ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

So, applying ideal gas equation for the remaining gas

(P-dP) V_o=(n-dn) RT

PV_o-dPV_o=nRT-dnRT …… (1)

Applying ideal gas equation, before gas was taken out

PV_o = nRT ……. (2)

Using equation (2) in (1) we get

nRT - V_odP = nRT - dnRT

V_odP=dnRT …… (3)

According to question, pressure of gas being taken out is equal to inner pressure of gas always. So inner pressure is equal to P-dP

Let the volume of gas taken out dV.

Applying ideal gas equation to gas pumped out

(P-dP) dV=dnRT

PdV=dnRT …… (4)

Where we have ignored dPdV as it is very small and can be neglected.

From equation (3) and (4)

V_odP=PdV

Given . Since volume is decreasing so rate should be negative.

Putting this value of dV in equation (5)

(a) Integrating equation (6) from Po to P and t=0 to t

Where we have used the formula

And

Taking exponential on both sides,

The pressure of the gas as a function of time is given as

(b) In second part the final pressure becomes half of the initial pressure

Putting this value of P in equation (7)

Taking natural logarithm on both side

The time taken before half the original gas is pumped out is

Given

Multiplying both sides by V

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant=8.31Jmol^-1K^-1

T=temperature

n=number of moles of gas

P=pressure of gas

Here, it is given that number of moles n=1

So, PV=RT. Putting this value of PV in equation 1

According to question V=V_o

The temperature of the gas when V = V₀ is .

We know that internal energy at a temperature

U=nC_vT

Where U=internal energy

n=number of moles

C_v=molar specific heat at constant volume

T=temperature

Air in home is chiefly diatomic molecules, so

C_v for diatomic moles is given as

So,

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant=8.31Jmol^-1K^-1

T=temperature

n=number of moles of gas

P=pressure of gas

Substituting the value of nRT from ideal gas equation to U

Now, it is given in question that pressure remains constant throughout the day and volume V of room is also constant. So, PV is a constant.

Hence U is also constant.

Given

Pressure of gas P₁=1atm=10⁵Pa

Radius of tube R=5cm=0.05m

So, area of tube = (0.05)²

Length of tube =20cm=0.2m

So, volume=arealength

Volume of cylindrical tube = (0.05)²0.2=0.0016m²

Initial temperature T₁ =300K

Final temperature T₂=600K

Coefficient of friction =0.2

Let final pressure be P₂. So, volume of the gas remains same until pressure becomes P₂ and then corks pop out. Number of moles will also be same. So, we can apply ideal gas equation which is

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

Net pressure on cork = P₂-P₁=210⁵ - 10⁵=10⁵Pa

We know that

So, force acting on the cork=pressure on corkarea of cork

F=10⁵(0.05)²

According to law of friction,

F=N

Where F=force of friction

N=normal to the surface of cork

=coffieicent of friction

Now, friction is always equal to the applied force until body starts to slide.

So,

In question N is denoted as dN. So, N=dN

And dl=length of cork around periphery of cork i.e. dl=circumference of cork.

Thus, the value of .

(a) Initially, the pressure inside the cylinder is equal to atmospheric pressure as given in question. So, pressure (thus force) on piston will be same from outside the cylinder and inside the cylinder which is atmospheric pressure. Therefore, no net force will act on piston and tension in wire will be zero.

(b)

Given

Initial temperature T₁=T_o

Increased temperature T₂=2T_o

Initial pressure of gas P₁=P_o=10⁵Pa

Let curved surface area of cylinder be A

Let the piston be L distance apart

Volume =arealength=AL

Since, no new gas is added to cylinder so, number of moles, before and after temperature change will be same and the volume of cylinder is same. So, applying ideal gas equation, before and after the increase of temperature.

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

P₂=210⁵=2P₁=2P_o

Net pressure P₂-P₁=2P_o-P_o=P_o

Net force acting outwards is force=pressurearea

F=P_oA ……(i)

Since, temperature is increased, gas inside the cylinder will expand and piston will move outward freely without any acceleration, as the piston is frictionless. So, from diagram F>T

From newton second law of motion, which states that ‘net external force on particle is equal to rate of change of momentum’, we can write

F-T=0

Rate of change of momentum will be zero because piston moves without any acceleration. So,

Therefore,

F=T= P_oA ( from (i))

The tension if the temperature is increased to 2T₀ is P_oA.

Given

Initial height of air trapped =h_o

Initial pressure=2P_o

P_o=atmospheric pressure = 10⁵Pa

Depth at which there is hole in tank =h₁

Let the density of water be

From the diagram we can see that,

Pressure of water above the water level of the bigger tank is given by

P=(h₂+h_o)g

Let the atmospheric pressure above the tube be P_o.

Total pressure above the tube=P_o+P=(h₂+h_o)g+P_o

This pressure initially is balanced by pressure above the tank 2P_o (from diagram).

Therefore,

2P_o=(h₂+h_o)g+P_o

P_o=(h₂-h₀)g

The height h₂ of the water in the long tube above the top initially is given by

(b)

Efflux means water flowing out from tube or outlet.

Velocity of efflux out of the outlet depends upon the total pressure above the outlet.

Total pressure above the outlet=2P_o+(h₁-h_o)g

Let the velocity of efflux be v₁ and the velocity with which the level of tank falls be v₂.

Pressure outside the outlet =P_o

Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy.

Mathematically Bernoulli’s theorem can be written as

Where v=velocity of fluid at a point

z= elevation of that point from a reference level

P=pressure of fluid at that point

=density of that fluid

G=acceleration due to gravity

Applying Bernoulli’s theorem, at points outside the outlet and above the outlet, we get

Consider the difference in elevation of both the points very small, so that we can ignore ‘gz’ term on both the sides.

Again, the speed with which the water level of the tank goes out is very less compared to the velocity of the efflux. Thus, v₂=0

The speed with which water comes out of the hole is given by

(c) Water maintains its own level, so height of the water of the tank will be h₁ when water will stop flowing out.

Thus, height of water in the tube below the tank height will be =h₁

Hence height if the water above the tank will be =-h₁

Given

Cross sectional area A =10cm²=1010^-4m²

mass of piston = 1kg

Pressure inside chamber = 100kPa=10⁵Pa

Pressure due to the weight of the piston =P_p

Pressure of vessel P₁=pressure of chamber+ pressure due to piston

P1=10⁵+9.810³

Volume of gas inside the vessel V₁=length of gas columnarea.

Given length of gas column=20cm=0.2m

V₁=length A

V₁=0.21010^-4=210^-4m³

After evacuation, pressure of chamber will be zero. So, pressure inside vessel after evacuation P2 will just be pressure due to piston.

P₂=0+9.810³

Let L be the final length of the gas column

Final volume V₂=AL=1010^-4L

Since, it is given in question that temperature remains constant, we can apply boyle’s law which states that ‘PV=constant when temperature is constant’.

Applying boyle’s law before and after evacuation,

P₁V₁=P₂V₂

(10⁵+9.810³)210^-4=9.810³1010^-4L

The length of the gas column after evacuation =2.2m

Given

Area of cross section A= 10cm²=1010^-4m²

Mass of piston ‘m’ = 1kg

Weight of piston ‘mg’=19.8N

Length of the gas column l=20cm=0.20m

Atmospheric pressure P_o=100kPa=10⁵Pa

Air pressure in the spaceship =100kPa=P_o

Let the length of the gas column in the spaceship be l’.

Pressure on gas before taking to spaceship= P₁

P₁= pressure due to weight of piston + atmospheric pressure

Now,

Volume of the gas column before taking to spaceship V1=arealength

V₁=Al

Volume of the gas after taking to spaceship V₂=Al’

The pressure of surroundings has been kept same as the atmospheric pressure. So, this means the temperature of surrounding, before and after taking to spaceship is same.

Therefore, we can apply boyle’s law we state that ‘PV=constant, when temperature is constant’.

We also know that in satellite or in spaceship the effect of gravity is negligible. So, pressure on gas due to weight of piston in spaceship will be zero.

So, pressure on the gas in spaceship P₂=air pressure of spaceship=P_o

Applying boyle’s law before and after taking vessel to spaceship, we get

P₁V₁=P₂V₂

(9.810³+10⁵)0.2=10⁵l’

109.810³0.2=10⁵l’

The length of gas column in spaceship is 22cm.

Given

Initial temperature of gas in both bulbs T₁ =0

Initial pressure of gas in both bulbs P₁=P₂=76cm of Hg=0.76m of Hg

Temperature of bulb placed in ice T₂=0=273.15K

Temperature of other bulb T’₂=62=335.15K

Let each of the bulbs have n₁ moles initially.

Now since the second bulb has kept at higher temperature gas in second bulb will expand and some of the gas will flow to first bulb and number of moles will change in second bulb.

Let the number of moles in second bulb after its pressure reached P be n₂.

Volume of both the bulbs is same so V₁=V₂=V.

Applying ideal gas equation in both bulbs

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

and equating the value of constant R

Number of moles left out of second bulb after temperature rose=n₁-n₂

Let n₃ moles be left when pressure reached P in fist bulb. Applying ideal gas equation in first bulb, before and after temperature change.

Also,

=own moles of first bulb n₁+moles received from second bulb

=n₁+n₁-n₂

Substituting the value of n₃ and n₂ in above equation

Taking n₁ common

P=0.8375m of Hg

The new value of the pressure inside the bulbs is 83.75cm of Hg.

Dew point is 20.

Explanation

Formula for relative humidity is given as

Where both actual vapor pressure and saturation vapor pressure are at same temperature.

Given

Temperature of air =20

Relative humidity=100%

So,

actual vapor pressure =saturation vapor pressure at the temperature 20

The temperature at which relative humidity is 100% i.e. air is completely saturated, is called dew point.

Here, the temperature at which the air is completely saturated is 20.

So, dew point is 20.

Given

Temperature T = 25=298K

Relative humidity =60%

Initial pressure of room=104kPa=1.04Pa

Saturation vapor pressure=3.2kPa=3.2Pa

Formula for relative humidity is given as

Where both actual vapor pressure and saturation vapor pressure are at same temperature.

So,

Therefore, actual vapor pressure of water vapor=0.6 saturation vapor pressure

Vapor pressure of water vapor=0.63.210³=1.9210³Pa

If all the water vapor is removed, then new pressure =pressure of room -pressure due to water vapours

=1.0410⁵-1.9210⁵

= 1.0210⁵Pa

If all the water vapour is removed from the room without changing the temperature, the new pressure will be 102kPa.

Given

Temperature of open room = 20°C

Dew point =10°C

New temperature of room =15 °C

The temperature at which relative humidity is 100% i.e. air is completely saturated, is called dew point.

So, dew point will not change until the temperature of room is less than dew point.

In our question, room temperature drops to 15 which greater than dew point.

So, the new dew point will be same as the old one.

Given

Relative humidity= 40%

Initial volume of vapor V₁=10cm³=10.

Let saturation vapor pressure be P_o

So, initial pressure of vapours= P₁=0.4P_o

As mentioned in question, final stage comes when vapours start to condense.

When vapor starts to condense, relative humidity becomes 100% and vapor pressure attains the maximum pressure called saturation vapor pressure.

So final pressure of vapor P₂=P_o

Since the process is isothermal, temperature remains constant throughout. Applying Boyle’s law which states that PV=constant, if temperature is constant.

P₁V₁=P₂V₂

the volume of the vapour at which it will start condensing is 410^-6m³

The reading of barometer gives the pressure in cm of Hg

Given

Reading of barometer on a day= atmospheric pressure of that day P= 76cm of Hg=0.76m of Hg

Drop in pressure when water is introduced P’=75.4cm of Hg

=0.754m of Hg

So, vapor pressure =P-P’=0.76-0.754=0.006m of Hg

Given saturation vapor pressure = 1cm of Hg=0.01m of Hg

So

the relative humidity in the space above the mercury column is 60%.

Boiling point of methyl alcohol at 1atm is 65 and at 0.5 atm is 48.

Explanation

To find the boiling point from above figure, we must look for the temperature on x-axis corresponding to pressures given in question.

So, for first pressure i.e. 1atm=760mm of Hg, the corresponding temperature on x- axis is 65. This shown by yellow line in figure.

For the second pressure i.e. 0.5 atm

1atm=760mm of Hg

0.5atm =7600.5 mm of Hg= 375mm of Hg

For 375mm of Hg the corresponding temperature on x-axis is 48. This is shown by orange line in the figure.

Minimum atmospheric pressure necessary to prevent blood from boiling is 50mm of Hg.

Explanation

Given

Average temperature of human body=98

T ()=98

T ()=

Now above 36.7, the human blood will start boiling. So, the minimum atmospheric pressure to prevent boiling of human blood will the pressure corresponding to this temperature.

So, from figure pressure corresponding to 36.7 temperature on y-axis is 50mm of Hg which shown by yellow line in the figure.

Given

Temperature of water=20

Droplets starts to form at 10.

Therefore, dew point is =10. This is because at dew point vapor pressure becomes saturated vapor pressure and after that air cannot hold more moisture and will start to condense.

At boiling point, saturation vapor pressure becomes equals to atmospheric pressure.

So, given that,

At temperature 20 pressure is 17.5mm of Hg.

SVP (saturation vapor pressure) at 20 = 17.5mm of Hg

SVP at dew point i.e. 10 =8.9mm of Hg

Now relation between relative humidity and dew point is

Relative humidity in the room is 51%.

Given

Initial temperature =30

Final temperature =20

Absolute humidity of saturated water vapor at 30= 30 g m^–3

This means 1m³ of air contains 30g of water vapor at 30.

So, amount of water vapor in 50m³ of air at 30=5030=1500g

Absolute humidity of saturated water vapor at 20= 16 g m^–3

So, amount of water vapor in 50m³ of air at 20=5016=800g

Amount water vapor condensed from 30 to 20 = 1500-800

= 700g

The mass of the water condensed is 700g.

Given:

Atmospheric pressure=76cm of Hg

Saturation vapor pressure=0.80 cm of Hg

When the water is introduced in barometer, water evaporates.

Thus, it exerts its vapor pressure over the mercury meniscus.

As more and more water evaporates, the vapor pressure increases that forces down the mercury level further down.

Finally, when the volume is saturated with the vapor at atmospheric temperature, the highest vapor pressure, i.e. saturation vapor pressure is observed, and the fall of mercury level reaches its minimum.

Thus,

Net pressure acting on the column = atmospheric pressure -saturation vapor pressure

= 76-0.80 =75.2 cm of Hg.

The height of the mercury column when it reaches its minimum value is 75.2cm.

Given

Volume of oxygen = 50cc=50cm³=50

Atmospheric pressure P_o=99.4kPa=99.4Pa

Temperature =27=300.15K

Saturation vapor pressure P_s=3.4kPa=3.4Pa

According to question water level in the jar is same as level outside. So

Pressure inside the jar = pressure outside the jar

Pressure outside the jar =atmospheric pressure P_o

Pressure inside the jar=P_o …. (1)

But

Pressure inside the jar is also = vapor pressure of oxygen+ Saturation vapor pressure

Pressure inside the jar=P+P_s …. (2)

From equation (1) and (2), we can write

P_o=P+P_s

P=P_o-P_s=99.4-3.4=96Pa

Applying ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

The number of moles of oxygen collected in the jar is 1.9310^-3.

Let the length of barometer be x cm.

Let the curved surface area of barometer tube be A

Given

Saturation vapor pressure (SVP) = 1.0 cm of mercury

In first case length of mercury column is 74cm.

Length of air above mercury = x-74

Volume of air column V₁= (x-74)A

In first case atmospheric pressure is 76 cm of Hg.

Let pressure of air column be P₁ in first case.

Then,

Atmospheric pressure= SVP+ P₁+mercury column height

76=1+P+74

P₁=1 …. (1)

In second case,

Length of mercury column is 72.10cm.

Length of air above mercury = x-72.10

Volume of air column V₂= (x-72.1)A

Atmospheric pressure is 74 cm of Hg.

Let pressure of air column be P_2.

Atmospheric pressure= SVP+ P₂ +mercury column height

74=1+P₂+72.1

74=P₂+73.1

P₂=74-73.1=0.9

P₂=0.9 …. (2)

Since temperature has not changed, we can apply Boyle’s law which states that PV=constant, if temperature is constant, for both the cases

P₁V₁=P₂V₂

1(x-74)A=0.9(x-72.1)A

(x-74)A=0.9(x-72.1)A

0.1x=9.11

x=91.1cm

Therefore, length of barometer tube is 91.1cm.

Given:

Temperature of air outside the room=0=273.15K

Temperature of air inside the room=20=293.15K

Relative humidity at 0 =40%

Let vapor pressure of air P₁.

SVP at 0=4.6mm of Hg

P₁=0.44.6=1.84mm of Hg

Volume of room is not changed so, applying ideal gas equation at 0 and 20.

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

We get,

Relative humidity inside home i.e. at temperature 20

SVP at 20=18mm of Hg

The relative humidity in the room is 10.9%.

Given

Temperature =27=300K

Relative humidity=50%

Volume =1m³

saturation vapor pressure (SVP) at 27°C = 3600 Pa

Vapor pressure P=0.53600=1800Pa

Molar mass of water M=16+2=18g

Let m₁ be the mass of water present in 50% humid air.

Applying ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

Number of moles n=

m₁=13g

Required pressure for saturation =3600Pa

Let m₂ be the amount of water required for saturation.

Again, applying ideal gas equation

Total excess water vapor that must be added=m₂-m₁=26-13=13g.

Given

Temperature T=300K

Relative humidity=20%

Volume of room= 50m³

Saturation vapor pressure at 300 K is 3.3 kPa=3.3Pa.

P=0.23.310³=660Pa

Molar mass of water H₂O =2+16=18g

Applying ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

Number of moles n=

m=238.55g

Mass of water vapor present in the room=238.55g.

Given

Temperature T=300K

Relative humidity=20%

Volume of room= 50m³

Mass of water=500g

Molar mass of H₂O M=2+16=18g

Saturation vapor pressure(SVP) at 300 K = 3.3 kPa=3300Pa

Let vapor pressure inside the room be P₁

P₁=0.23.310³=660Pa

Since the floor has dried that means water on the floor has been evaporated.

Let P₂ be the partial pressure of evaporated water

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

And Number of moles n=

P₂=1385Pa

Total pressure of room= partial pressure of evaporated water+ pressure of air inside the room

P=P₂+P₁

P=1385+660=2045Pa

The relative humidity when the floor dries is 61.9%.

Given

Temperature =15

Relative humidity=40%

Volume =50cm³

Molar mass of H₂O M=2+16=18g.

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^-1mol^-1

T=temperature

n=number of moles of gas

P=pressure of gas.

The saturation vapor pressure (SVP)of water at 15°C=1.6kPa=1.610³Pa.

P=0.41.610³Pa

Evaporation will occur if the atmosphere is not saturated.

Net pressure change= SVP-P

P’ =1.6103-0.41.610³=0.9610³Pa

Number of moles n=

Applying equation of ideal gas.

The amount of water that will evaporate=361g

(b)

SVP at 20=2.4kPa

SVP at 15=1.6kPa

Net pressure changes in increasing the temperature,

Net pressure changes P’’=2.4-1.6=0.8kPa=0.8Pa.

Applying equation of ideal gas.

So, if the room temperature is increased by 5°C, the amount of water will evaporate is 296g.