ROUTERA


Specific Heat Capacities Of Gases

Class 12th Concepts Of Physics Part 2 HC Verma Solution



Short Answer
Question 1.

Does gas have just two specific heat capacities or more than two? Is the number of specific heat capacities of a gas countable?


Answer:

No, the number of specific heat capacities of gas is infinite as it depends on the thermodynamic process followed by the gas.


The specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of one mole of that substance by 1 degree Celsius, or 1 Kelvin. It is denoted by C.


Gases are compressible substances. They have two well-known specific heat capacities: one at constant pressure(Cp)(isobaric process - constant pressure) and another at constant volume(Cv)(isochoric process - constant volume). However, gases can have many specific heat capacities depending on the other thermodynamic processes they follow, like adiabatic process, isothermal process, etc.



Question 2.

Can we define specific heat capacity at constant temperature?


Answer:

The specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of one mole of that substance by 1 degree Celsius, or 1 Kelvin. It is denoted by C.


Hence, , … (i)


where


C = specific heat capacity


Q = heat required to raise the temperature by dT


m = molar mass


dT = change in temperature.


For constant temperature, dT = 0. Putting this value in (i), we get



Hence, for a process at a constant temperature, the specific heat capacity is infinite.



Question 3.

Can we define specific heat capacity for an adiabatic process?


Answer:

The specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of one mole of that substance by 1 degree Celsius, or 1 Kelvin. It is denoted by C.


Hence, , … (i) where


C = specific heat capacity


Q = heat required to raise the temperature by dT


m = molar mass


dT = change in temperature.


For an adiabatic process, Q = 0. Substituting this value in (i), we get


C = 0.


Hence, for a process at a constant temperature, the specific heat capacity is zero.



Question 4.

Does a solid also have two kinds of molar heat capacities CP and CV? If yes, do we have CP > CV? Is CP – CV = R?


Answer:

Yes, a solid also has two kinds of molar heat capacities, Cp(at constant pressure) and Cv(at constant volume).


Solids are almost incompressible.


Hence, the values of Cp and Cv are such that Cp > Cv, but they are almost equal since the dependence on heat capacities is very less in the case of solids.


Since the values of Cp and Cv are not that different, Cp - Cv is much less than R.



Question 5.

In a real gas the internal energy depends on temperature and also on volume. The energy increases when the gas expands isothermally. Looking into the derivation of CP – CV = R, find whether CP – Cv will be more than R, less than R or equal to R for a real gas.


Answer:

We know that, for an ideal gas,


CP - CV = R … (i)


where


Cp = specific heat constant at constant pressure


Cv = specific heat constant at constant volume


R = universal gas constant


Multiplying by n x dT on both sides of (i), we get


nCPdT - nCvdT = nRdT


which gives


(dQ)P - (dQ)v = nRdT …(ii)


Since (dQ)p = nCpdT and (dQ)v = nCvdT … (iii)


Where


n = number of moles


dT = change in temperature


(dQ)p = change in heat at constant pressure


(dQ)v = change in heat at constant volume


However, for a real gas, the internal energy depends on the temperature as well as the volume.


Hence, there will be an additional term on the right-hand side of (ii) which will indicate the change in the internal energy of the gas with volume at constant pressure. Let this term be u.


Hence, for a real gas, (ii) becomes :


(dQ)p - (dQ)v = nRdT + u … (iv)


Again, dividing on both sides by ndT, we get


… (v),


which is greater than R.


Here,


Cp = specific heat constant at constant pressure


Cv = specific heat constant at constant volume


R = universal gas constant


n = number of moles


dT = change in temperature


Hence, from (v), we get Cp - Cv > R.


We conclude that for a real gas, Cp - Cv > R.



Question 6.

Can a process on an ideal gas be both adiabatic and isothermal?


Answer:

According to the first law of thermodynamics,


dQ = dU + dW = nCvdT + dW … (i),


where


dQ = heat supplied


dU = change in internal energy


dW = work done on the gas


n= number of moles


Cv = specific heat capacity at constant volume


dT = change in temperature


For an adiabatic process, dQ(heat supplied) = 0.


An adiabatic process occurs without the transfer of heat or mass of substances between the thermodynamic system and the surrounding.


For an isothermal process, dT(change in temperature) = 0


An Isothermal process is a change of system, in which the temperature remains constant ∆T=0.


Putting these values in (i), we get


dW = 0,


which is not possible for either of the processes.


dW = 0 only in the case of a process where the volume is constant that is dV = 0,


since dW = PdV,


where P = pressure and dV = change in volume.


Hence, we conclude that a process cannot be both adiabatic and isothermal.



Question 7.

Show that the slope of p –V diagram is greater for an adiabatic process as compared to an isothermal process.


Answer:

For an isothermal process, the ideal gas equation is given as


PV = constant … (i),


Where


P = pressure


V = volume.


Differentiating on both sides of (i), we get


PdV + VdP = 0


On solving for , we get


… (ii)


For a graph of P versus V, dP/dV indicates the slope.


Hence, for an isothermal process, the slope of the p-V diagram is given by -P/V.


Now for an adiabatic process, the ideal gas equation is


PVγ= constant … (iii),


where


P = pressure,


V = volume,


γ = ratio of specific heat capacities at constant pressure and constant volume.


Differentiating both sides of (ii), we get


V𝛾dP + 𝛾V𝛾-1PdV = 0 which gives


… (iv)


Hence, for an adiabatic process, the slope of the p-V diagram is given by -𝛾P/V.


Since 𝛾 > 1, we find that 𝛾P/V is greater than P/V, which concludes that slope of p-V diagram of an adiabatic process is steeper than that of an isothermal process(proved).



Question 8.

Is a slow process always isothermal? Is a quick process always adiabatic?


Answer:

An isothermal process is represented by the equation


PV = constant … (i),


where


P = pressure


V = volume.


To keep this product constant, a small change in V will only produce a small change in P and vice versa. Hence, an isothermal process is usually a slow process.


On the other hand, an adiabatic process is represented as


PV𝛾 = constant … (ii), where


P = pressure


V = volume


The 𝛾= ratio of specific heat capacities at constant pressure to constant volume.


Now, 𝛾 > 1. Hence, the term V𝛾 will increase exponentially. Hence, to keep the product constant, a small change in V will cause a large change in P. Hence, an adiabatic process is usually a fast process.



Question 9.

Can two states of an ideal gas be connected by an isothermal process as well as an adiabatic process?


Answer:

An isothermal process is represented by


PV = constant … (i),


where P = pressure, V = volume.


This gives


P1 V1 = P2 V2 (ii),


Where


P1, V1 = initial values of pressure and volume


P2, V2 = final values of pressure and volume.


An adiabatic process is represented by


PV𝛾 = constant… (iii),


where


P = pressure,


V = volume,


𝛾 = ratio of specific heat capacities at constant pressure and constant volume.


From (iii), we get


P1V1𝛾 = P2V2𝛾… (iv),


Where P1, V1 = initial values of pressure and volume


P2, V2 = final values of pressure and volume.


Dividing (iv) by (ii), we get


V1𝛾-1 = V2𝛾-1 … (v)


If the condition given by (v) is satisfied, then two states of an ideal gas can be connected by both an isothermal and an adiabatic process.



Question 10.

The ratio CP/CV for gas is 1.29. What is the degree of freedom of the molecules of this gas?


Answer:

Given:


= 1.29


Formula used:


… (i),


where


γ = ratio of molar heat capacities at constant pressure to constant volume


f = number of degrees of freedom


(i) becomes :



=> … (ii)


Substituting γ = 1.29 in (ii), we get


f = = 6.89 which is approximately equal to 7.


Thus, the number of degrees of freedom is approximately equal to 7.




Objective I
Question 1.

Work done by a sample of an ideal gas in a process A is double the work done in another process B. The temperature rises through the same amount in the two processes. If CA and CB be the molar heat capacities for the two processes.
A. CA = CB

B. CA < CB

C. CA > CB

D. CA and CB cannot be defined


Answer:

Q = nCdT … (i),


where Q = work done by an ideal gas


n = number of moles


C = molar heat capacity


dT = rise in temperature.


For process A, let the value of C be CA and for B, let it be CB.


Since the work done by the gas in process A is twice that in B, and the rise in temperature is the same in both the cases, we get two equations:


2Q = nCAdT … (ii) and


Q = nCBdT … (iii),


Where


Q = work done in process B


n = number of moles of gas


dT = rise in temperature


Dividing (ii) by (iii), we get



=> CA = 2CB


This proves that CA>CB.


Question 2.

For a solid with a small expansion coefficient,
A. CP – CV = R

B. CP = CV

C. CP is slightly greater than CV

D. CP is slightly less than CV


Answer:

For solids which have a small expansion coefficient, the work done on the solid is pretty small. Hence, the specific heat at constant pressure and at a constant volume only slightly different since the work done depends very little on the process. Hence, Cp is slightly greater than Cv but much less than R as in the case of ideal gases.


Question 3.

The value of CP – CV is 1.00 R for a gas sample in state A and is 1.08R in state B. Let PA, PB denote the pressures and TA, and TB denote the temperatures of the states A and B respectively. Most likely
A. pA < pB and TA > TB

B. pA > pB and TA < TB

C. pA = pB and TA < TB

D. pA > pB and TA = TB


Answer:

For gas in state A, Cp - Cv = R. This represents an ideal gas. Now, for an ideal gas, we require very high temperature and very low pressure compared to a real gas


For gas in state B, Cp - Cv = 1.08R, which represents a real gas. Since gas A was ideal, its pressure must be much lower than that of B and temperature must be much higher than that of A.


Hence, we require the condition pA < pB and TA > TB. This is given by option (a).


Options (b), (c) and (d) are incorrect because none of those satisfies the conditions for an ideal gas.


Question 4.

Let CV and CP denote the molar heat capacities of an ideal gas at constant volume and constant pressure respectively. Which of the following is a universal constant?
A. CP/CV

B. CP CV

C. CP – CV

D. CP +CV


Answer:

For an ideal gas, Cp - Cv = R.


Here, R is the universal gas constant.


Hence, the correct option is (c).


Option (a) is incorrect since Cp/Cv = ℽ, which differs for monoatomic, diatomic or polyatomic gases. Here ℽ = ratio of molar heat capacities at constant pressure and constant volume.


Option (b) is incorrect since CpCv = ℽCv2, which is not a constant as ℽ varies.


Option (d) is incorrect since Cp + Cv = (ℽ+1)Cv, which is not a constant as ℽ varies.


Question 5.

70 calories of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from 30°C to 35°C. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is



A. 30 calories

B. 50 calories

C. 70 calories

D. 90 calories


Answer:

We know that, Q = nCpdT… (i),


Where


Q = heat required to raise the temperature


n = number of moles


Cp = specific heat capacity at constant pressure


dT = rise in temperature.


In this first case,


Amount of heat required(Q) = 70 cal


Number of moles(n) = 2 mol


Rise in temperature(dT) = (35-30)0C = 50C


Hence, from (i), we get Cp = 7 calmol-1 0C-1


Now, we know, Cp - Cv = R,


Where


Cp = specific heat capacity at constant pressure


Cv = specific heat capacity at constant volume


R = universal gas constant = 1.98 calmol-1 0C-1


Therefore, we get Cv = Cp - R = (7-1.98) cal mol-1 0C-1


= 5.02 cal mol-1 0C-1


Now, in the new case, change in temperature dT = 50C as before. Number of moles(n) = 2, and Cv = 5.02 calmol-1 0C-1


Hence, amount of heat required to raise the temperature of the same gas through the same range


dQ = nCvdT = (2 x 5.02 x 5) cal = 50.2 cal which is approximately equal to 50 cal.


Hence, the correct option is (b).


Question 6.

The molar heat capacity for the process shown in the figure is




A. C = Cp
B. C = Cv
C. C > Cv
D. C = 0



Answer:

Let us consider the figure given below :



We consider the process AB to be composed of two processes, AC(at constant pressure) and CB(at constant volume), such that AB = AC + CB. C is the molar heat capacity of AB, Cp is the molar heat capacity of AC(constant pressure) and Cv is the molar heat capacity of CB(constant volume).


From the first law of thermodynamics, we know that Q = U + W … (i),


Where Q = heat supplied, U = change in internal energy, W = work done on the system.


Since the change in internal energy is independent of the path, it will have the same value for paths AB and ACB.


Hence, UAB = UACB … (ii), where UAB = change in internal energy for path AB, UACB = change in internal energy for path ACB.


Now, the work done by a process is given by the area under the PV diagram. We can clearly see that the area under process AB is greater than the sum of areas under processes AC and CB.


Hence, WAB > WACB … (iii), where WAB = work done for process following path AB, WACB = work done for process following path ACB.


Adding (ii) and (iii), we get :


UAB + WAB > UACB + WACB … (iv)


But, from the first law of thermodynamics, Q = U + W, where Q = heat supplied, U = change in internal energy, W = work done on the system.


Hence, we get, QAB > QACB … (v),


We know, QAB = nCdT, QAC = nCpdT, QCB = nCvdT,


where QAB = heat supplied in process AB


QAC = heat supplied in process AC


QCB = heat supplied in process CB


n = number of moles


C = molar heat capacity


Cp = molar heat capacity at constant pressure


Cv = molar heat capacity at constant volume


dT = change in temperature.


Also, QACB = QAC + QCB … (vi), where


QACB = heat supplied in process ACB


QAC = heat supplied in process AC


QCB = heat supplied in process CB


Hence, from (v), we get,


nCdT > nCpdT + nCvdT


Dividing by ndT on both sides :


C > Cp + Cv => C > Cv (option c)


Question 7.

The molar heat capacity for the process shown in the figure is



A. C = CP

B. C = CV

C. C > CV

D. C = 0


Answer:

Given : which gives us


pVg = K, where p = pressure, V = volume, and g and K are constants. This represents an adiabatic process.


In an adiabatic process, Q(heat exchanged) = 0.


Now we know that Q = nCdT,


where n = number of moles, C = specific molar heat capacity and dT = rise in temperature.


Since Q = 0, we get C = 0.


Hence, the correct option is (d).


Question 8.

In an isothermal process on an ideal gas, the pressure increases by 0.5%. The volume decreases by about
A. 0.25%

B. 0.5%

C. 0.7 %

D. 1%


Answer:

Let the initial values of pressure and volume be P1 and V1, and let the final values of pressure and volume be P2 and V2.


Now for an isothermal process, PV = constant .. (i)


This gives us: P1V1 = P2V2.. (ii)


Since the pressure increases by 0.5%, the new pressure is



Substituting this value in (ii) :


P1V1 = 1.005P1V2


=> .


Hence, the volume decreases by (1-0.995) = 0.005 = 0.5%


Hence, the correct option is (b).


Question 9.

In an adiabatic process on gas with γ = 1.4, the pressure is increased by 0.5%. The volume decreases by about
A. 0.36%

B. 0.5%

C. 0.7%

D. 1%


Answer:

For an adiabatic process, PV𝛾 = constant … (i)


If P1, V1 represents the initial values of pressure and volume and P2, V2 represent the final values of pressure and volume,


P1V1𝛾 = P2V2𝛾… (ii)


Now, 𝛾 = 1.4 (given).


It is given that the pressure increases by 0.5%.


Hence, .


Substituting in (ii) :


P1V1𝛾 = 1.005 X P1V2𝛾


=> (V1/V2)γ = 1.005 … (iii)


Taking log on both sides of (iii), we get



… (iv)


Taking inverse log on both sides of (iv), we get


V1/V2 = 100.0014 = 1.003


=> V2 = V1/1.003 = 0.997V1


Hence, the volume decreases by (1-0.997) = 0.003 which is approximately equal to 0.36%.


Hence, the correct option is (a).


Question 10.

Two samples A and B are initially kept in the same state. Sample A is expanded through an adiabatic process and sample B through an isothermal process. The final volumes of the samples are the same. The final pressures in A and B are pA and PB respectively.
A. PA > PB

B. PA = PB

C. PA < PB

D. The relation between pA and PB cannot be deduced.


Answer:

Adiabatic process is represented by : PV𝛾 = constant,


where P = pressure


V = volume


ℽ = ratio of specific heat capacities at constant pressure and constant volume


Therefore, we get PA = constant/V. .. (i)


Isothermal process is given by : PV = constant


Hence, we get PB = constant/V … (ii)


Since ℽ > 1, we get pA < pB. (proved)


Question 11.

Let Ta and Tb be the final temperatures of the samples A and B respectively in the previous question.
A. Ta < Tb

B. Ta = Tb

C. Ta > Tb

D. The relation between Ta and Tb cannot be deduced.


Answer:

Since sample B undergoes isothermal expansion, its temperature remains constant = Tb.


For an adiabatic process, since the heat supplied is 0, the internal energy will change by an amount dU = nCvdT,


where dU = change in internal energy


n = number of moles


Cv = specific heat capacity at constant volume


dT = change in temperature


This change in internal energy will compensate for the constancy in heat.


Sample B is undergoing expansion through an isothermal process; its initial and final temperatures will be the same.


Sample A will expand at the cost of its internal energy.


Therefore, the final temperature will be less than the initial temperature,


since dU < 0 => dT < 0.


Tb-Ta<0


Tb>Ta or Ta<Tb


Hence, we get Ta>Tb.


Question 12.

Let ΔWa and ΔWb be the work done by the systems A and B respectively in the previous question.
A. ΔWa > ΔWb

B. ΔWa = ΔWb

C. ΔWa < ΔWb

D. The relation between ΔWa and ΔWb cannot be deduced.


Answer:

We know that for any given state, the slope of the p-V diagram of an adiabatic process is -γP/V, while that of an isothermal process is -P/V.


Hence, the slope of an adiabatic process is more.


Now, the area under the curve of a p-V diagram gives the work done.



From this diagram, we can see that that area under the curve of process A(ACDE) which represents the adiabatic process with greater slope is less than that of the area under the curve of process B(ABDE), which represents the isotherm.


Hence, we can conclude that ΔWa < ΔWb


Question 13.

The molar heat capacity of oxygen gas at STP is nearly 2.5R. As the temperature is increased, it gradually increases and approaches 3.5R. The most appropriate reason for this behaviour is that at high temperatures.
A. oxygen does not behave as an ideal gas

B. oxygen molecules dissociate in atoms

C. the molecules collide more frequently

D. molecular vibrations gradually become effective.


Answer:

We know that the molar heat capacity of any gas depends on the degrees of freedom of the gas. On increasing the temperature of the gas, we allow more molecules to vibrate about their equilibrium positions, which in turn increases the degrees of freedom of the gas. For this reason, the molar heat capacity increases as the temperature of the gas are increased.


Hence, the correct answer is an option (d).



Objective Ii
Question 1.

A gas kept in a container of finite conductivity is suddenly compressed. The process
A. must be very nearly adiabatic

B. must be very nearly isothermal

C. may be very nearly adiabatic

D. may be very nearly isothermal


Answer:

Since the gas kept in the container is suddenly compressed, there is very little time for any heat to allow to enter or leave the container. Hence, it must be very nearly an adiabatic process. If the temperature remains constant during this process, then the process may also be isothermal. Option (a) is incorrect because it must be very nearly adiabatic but might not be completely depending on how


sudden the process is.


Option (b) is incorrect because if the compression is indeed done suddenly, then the temperature of the gas does not remain constant, and it may not be isothermal. It changes with the pressure and the volume of the gas according to the formulae:


i.


ii. PVᵞ = constant


Where,


P1, V1, T1: initial values of pressure, volume and temperature


P2, V2, T2: final values of pressure, volume and temperature


𝛄: adiabatic index, varies from gas to gas


Therefore, the correct answers are option (c) and (d).


Question 2.

Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an isothermal process.
A. Q = 0

B. W = 0

C. Q ≠ W

D. Q = W


Answer:

In an isothermal process, the temperature remains constant. The internal energy of an ideal gas is a state function that depends on temperature. Hence, change in internal energy is zero and from the first law of thermodynamics: ΔU = Q - W, where ΔU = change in internal energy, Q = amount of heat given and W = work done by it.


Since ΔU = 0 in this case, we get Q = W.


Options (a) and (b) are incorrect because we actually provide a finite amount of heat to the system, and hence work is also not zero.


Option (c) is incorrect because we just showed that Q = W.


Hence, the correct option is option (d).


Question 3.

Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an adiabatic process.
A. Q = 0

B. W = 0

C. Q = W

D. Q ≠ W


Answer:

In an adiabatic process, we assume that there is no heat exchange between the system and surroundings.


Hence, Q = 0. Also, U = W(when Q = 0), so Q ≠ W holds.


Option (b) is incorrect because due to change in temperature, there is still a change in internal energy and from the first law of thermodynamics, ΔU = W. Hence W ≠ 0.


Option (c) is incorrect because the heat exchange is zero, but the work done in general is not zero.


Hence, the correct answers are option (a) and (d).


Question 4.

Consider the processes A and B shown in figure. It is possible that



A. both the processes are isothermal

B. both the processes are adiabatic

C. A is isothermal and B is adiabatic

D. A is adiabatic and B is isothermal


Answer:

From the graph, we can see that the slope of process B is steeper than that of process A.


Now in an isothermal process, under constant temperature, Pressure(P) x Volume(V) = constant … (i) (according to Boyle’s law)


Differentiating on both sides:


PdV + VdP = 0.


slope = = … (ii)


For an adiabatic process,


PVᵞ = constant


Where,


𝛾 is the ratio of specific heat of the gas at constant pressure and constant volume.


Differentiating the above question, we get


VᵞdP + 𝛾V𝛾-1PdV= 0



which is greater than that in case of the isothermal process.


Thus path (A) is for isothermal process while path (B) is for the adiabatic process.


Options (a) and (b) are incorrect since the slopes of the two paths Differ.


Option (d) is incorrect because the slope of path (A) is less than that of path (B).


Hence, the correct answer is option (c).


Question 5.

Three identical adiabatic containers A, B and C contain helium, neon and oxygen respectively at equal pressure. The gases are pushed to half their original volumes.
A. The final temperatures in the three containers will be the same.

B. The final pressures in the three containers will be the same.

C. The pressures of helium and neon will be the same but that of oxygen will be different.

D. The temperatures of helium and neon will be the same but that of oxygen will be different.


Answer:

Since this is an adiabatic process, Q(heat) = 0.


Hence, dU(change in internal energy) = dW(work done) = nCvdT.


where n = no. of moles, Cv = specific heat of the gas at constant volume, and dT = change in temperature.


Cv is different for oxygen(diatomic gas) but same for helium and neon(monoatomic gases) so dT for oxygen is different.


W for adiabatic process = (P2V2-P1V1)/(1-𝛾)


In all cases, V2 = V1/2.


P1, V1, T1 - initial pressure, volume and temperature


P2, V2, T2 - final pressure, volume and temperature


V1, P1 are same for all gases and so is W. Only 𝛾 for oxygen is different compared to helium and neon since it is a diatomic gas while the rest are monoatomic.


Hence, options (c) and (d) are correct.


Question 6.

A rigid container of negligible heat capacity contains one mole of an ideal gas. The temperature of the gas increases by 1°C it 3.0 cal of heat is added to it. The gas may be
A. helium

B. argon

C. oxygen

D. carbon dioxide


Answer:

Let us use the formula,


Heat supplied to a gas, dQ = nCvdT.


Where,


n is the number of mole of the ags


Cv is the specific heat of the gas at constant volume T is the temperature


Putting the values in the above formula, we get


dQ = 3.0 cal = (3.0 x 4.2)J = 12.6 J.


dT = 1°C, n = 1.


Therefore, Cv is approximately equal to 12.6 J/mol/°C ~ 1.5R = 3R/2


We know,


Where Cp = specific heat at constant pressure


R = universal gas constant = 8.314 J/kg/mol


which holds for monoatomic gases.


Helium and Argon are monoatomic and so options (a) and (b) hold.


Question 7.

Four cylinders contain equal number of moles of argon, hydrogen, nitrogen and carbon dioxide at the same temperature. The energy is minimum in
A. argon

B. hydrogen

C. nitrogen

D. carbon dioxide


Answer:

According to the law of equipartition of energy, the Kinetic energy associated with each degree of freedom of a molecule is kT/2,


KE=


where k is Boltzmann constant,


T is absolute temperature.


f is the number of degrees of freedom


The number of degrees of freedom of a system is the minimum number of coordinates required to completely describe its position and orientation.


Now argon,


which is a monoatomic gas, can only translate along the x, y and z axes.


Hence, it has 3 degrees of freedom. Now, for a system,


total associated energy is:


Total energy of all the degrees of freedom,


Where


E = total energy,


f = number of degrees of freedom


k = Boltzmann constant = 1.38 x 10-23m2 kg s-2 K-1


T = absolute temperature (Kelvin)


For argon, f = 3. Therefore, which is minimum.


Options (b) and (c) are incorrect since they are diatomic and


have more than 3 degrees of freedom(translational and rotational degrees of freedom of 2 molecules).


Option (d) is incorrect since carbon dioxide is triatomic and has


more than 3 degrees of freedom(translational and rotational degrees of freedom of 3 molecules).


Hence, the correct answer is option (a).



Exercises
Question 1.

A vessel containing one mole of a monatomic ideal gas (molecular weight = 20 g mol–1) is moving on a floor at a speed of 50 ms–1. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.


Answer:

Given:


number of moles, n = 1


Specific heat at constant temperature, Cv(monoatomic gas) = 3R/2 = 12.471 J/mol/K


R = universal gas constant = 8.314 J/mol/K


Initial velocity(vi) = 50 m/s


Final velocity(vf) = 0


Molecular weight(m) = 20 g/mol = 0.02 kg/mol


Formula Used:


i. Change in internal energy(dU) = nCvdT,


Where,


n is the number of the moles of the gas,


Cv is the heat capacity at the constant volume


dT = rise in temperature.


ii. Mechanical energy lost,


Where,


m is the molecular weight of the gas in kg


vi is the initial velocity,


vf is the final velocity,


equating equation (i) and (ii), we get


Putting the values in the above equation, we get



=> dT = rise in temperature ~ 2 K (Answer).



Question 2.

5g of a gas is contained in a rigid container and is heated from 15°C to 25°C. Specific heat capacity of the gas at constant volume is 0.172 cal g–1°C–1 and the mechanical equivalent of heat is 4.2 J cal–1. Calculate the change in the internal energy of the gas.


Answer:

Given:


Mass of the gas(m) = 5g


Change in temperature(dT) = 10°C


Specific heat at constant volume(Cv) = = 0.172 cal g–1°C–1


Mechanical equivalent of heat = 4.2 J cal–1


Formula used:


Change in internal energy = mCvdT


Where,


m = Mass of the gas


Cv = Specific heat at constant volume


dT = Change in temperature


Also, we know,


Heat(Joule)= Mechanical equivalent of heat× Heat(cal)


Putting the values in the above formula, we get


= (5x10x0.172) cal


= 8.6 cal = (8.6x4.2)J


= 36.12 J (Answer)



Question 3.

Figure shows a cylindrical container containing oxygen (γ = 1.4) and closed by a 50 kg frictionless piston. The area of cross section is 100 cm2, atmospheric pressure is 100 kPa and g is 10 ms–2. The cylinder is slowly heated for some time. Find the amount of heat supplied to the gas if the piston moves out through a distance of 20 cm.




Answer:

Given:


Mass of piston(m) = 50 kg


Area(A) = 100 cm2 = (100x10-4)m2 (since 1m = 100 cm)


Acceleration due to gravity, g = 10 ms-2


Atmospheric pressure = 100 kPa


Distance through which it moves = 20 cm


γ = 1.4


Formula used:


Therefore, pressure exerted by piston = =


= ((50x10)/(100x10-4)) Pa


= 50,000 Pa


Atmospheric pressure = 100 kPa = 1,00,000 Pa.


Therefore, Total pressure(P) = (50,000 + 1,00,000)Pa


=1,50,000Pa


Work done = Pressure x change in volume = P x dV


dV(change in volume) = distance moved by piston x Area


= (20cm x 100cm2)


= 2,000 cm3 = 2,000 x 10-6 m3 = 2 x 10-3 m3


Therefore, Work = (1,50,000 x 2 x 10-3) J = 300 J


Work done, W= P∆V =n R dT


We get,



Now, We calculate Q:


dQ= nCpdt =


Given: γ = 1.4 = . Also,. Solving these two equations, we get Cp = 7R/2, Cv = 5R/2.


Hence, dQ = = 1050 J.



Question 4.

The specific heat capacities of hydrogen at constant volume and at constant pressure are 2.4 cal g−1 °C−1 and 3.4 cal g−1 °C−1 respectively. The molecular weight of hydrogen is 2 g mol−1 and the gas constant, R = 8.3 × 107 erg °C−1mol−1. Calculate the value of J


Answer:

Given:


Specific heat at constant volume is Cv =2.4 cal g-1oC-1


Specific heat at constant pressure is Cp=3.4 cal g-1oC-1


Molecular mass of hydrogen is 2 g mol-1


Gas constant, R=8.3× 107 gmol


Formula used:



Where:


m is the molecular weight of hydrogen


Cp is the specific heat at constant pressure


Cv is the specific heat at constant volume


R is the gas constant.


Putting the values in the above equation , we get


The gas constant R= (m X Cp ) - (m X Cv) = 2 × (Cp-Cv)= 2×J


Now, 2× J=R


2× J=8.3 × 107 erg/ mol-oC


Thus, J=4.15 × 107 erg/cal (Answer).



Question 5.

The ratio of the molar heat capacities of an ideal gas is CP/CV = 7/6. Calculate the change in internal energy of 1.0 mole of the gas when its temperature is raised by 50 K

(a) keeping the pressure constant,

(b) keeping the volume constant and

(c) adiabatically.


Answer:

Given:


n = number of moles = 1,


Cv = specific heat capacity at constant volume,


Cp = specific heat capacity at constant pressure


dT = change in temperature = 50K.


γ= Ratio of molar heat capacities = CP/CV = 7/6 => Cv = 6Cp/7.


(a) Formula used:


Pressure constant: Isobaric process. For an isobaric process,


change in internal energy dU = nCvdT,


Where


n = number of moles,


Cv = specific heat at constant volume,


dT = rise in temperature


Also, Cp-Cv = R.


Cp = specific heat at constant pressure


Cv = specific heat at constant volume


R = universal gas constant = 8.314 J/mol/K


Substituting: Cp - 6Cp/7 = Cp/7 = R => Cp = 7R.


Therefore Cv = Cp - R = 6R = (6 X 8.314)J/mol/K


Therefore,


dU = 1 mol X (6 X 8.314)J/mol/K X 50K = 2494.2 J(Ans)


(b) Volume constant: Isochoric process, dV = 0(change in volume)


First law of thermodynamics gives us: dU = dQ - dW


Where dU = change in internal energy, dQ = change in heat,


dW = work done = Pressure x change in volume = PdV


Since dV = 0, dU = dQ.


Hence, dU = nCvdT since dQ = nCvdT.


Where


n = Number of moles,


Cv = Specific heat at constant volume,


dT =Change in temperature


Putting the values in the above formula, we get


Therefore, dU = 1 mol x (6 x 8.314) J/mol/K x 50K


= 2494.2 J(Ans)


(c) Adiabatic process: dQ(heat change) = 0. Therefore,


dU(change in internal energy) = dW(work done)


Since dQ = dU + dW.


For an adiabatic process, dW = dT/(𝛾-1).


dW = work done, dT = change in temperature


𝛾 = Cp/Cv = 7/6 , Cp = specific heat capacity at constant


pressure, Cv = specific heat capacity at constant volume


Therefore dU = - dW = - = (8.314X50)/(7/6 - 1)


= 2494.2 J (Ans)



Question 6.

A sample of air weighing 1.18g occupies 1.0 × 103 cm3 when kept at 300K and 1.0 × 105 Pa. When 2.0 cal of heat is added to it at constant volume, its temperature increases by 1°C. Calculate the amount of heat needed to increase the temperature of air by 1°C at constant pressure if the mechanical equivalent of heat is 4.2 × 107 erg cal–1. Assume that air behaves as an ideal gas.


Answer:

Given:


Pressure, p= 1.0 × 105 Pa,


Temperature, T = 300K,


Universal gas constant, R = 8.314 J/kg/mol


V(volume) = 1.0 × 103 cm3 = 0.001 m3


Formula used:


1. Ideal gas equation: PV = nRT.


Where,


P = pressure,


V = volume,


n = number of moles,


R = universal gas constant = 8.314 J/kg/mol,


T = absolute temperature


2. Number of moles n = PV/RT = 100000x0.001/(8.314*300) = 0.04 mol


3. First law of thermodynamics: dQ = dU + dW = dU + PdV,


Where,


dQ = heat supplied,


dU = change in internal energy,


dW = PdV = work done, where P = pressure, dV = change in volume.


Since volume is constant, dV = 0 => dW = 0.


Hence, dQ = dU.


Heat(dQ) = 2 cal = nCvdT = 0.04 mol x Cv x 1K ,


Where n = number of moles, Cv = specific heat at constant volume, dT = rise in temperature.


=> Cv = 50 cal/mol/K. = (50 x 4.2 × 107)erg/cal x cal/mol/K = 2.1 x 109 erg/mol/K = 210 J/mol/K,


Since heat(J) = mechanical equivalent of heat x heat(cal) = 4.2 x heat(cal), and


1 J = 107 erg


We know, Cp = (Cv+R),


Where Cp = specific heat at constant pressure, Cv = specific heat at constant volume, R = universal gas constant = 8.314 J/kg/mol


Therefore, Cp = (210+8.314) J/mol/K = 218.314 J/mol/K.


Therefore, heat required to raise the temperature by 1°C at constant pressure = nCpdT,


Where n = number of moles, Cp = specific heat at constant pressure, dT = rise in temperature.


Hence, substituting, heat = (0.04 x 218.314 x 1) J = 8.737 J = (8.737/4.2) cal = 2.08 cal (since 1 J = 4.2 cal)(Ans)



Question 7.

An ideal gas expands from 100 cm3 to 200 cm3 at a constant pressure of 2.0 × 105 Pa when 50J of heat is supplied to it. Calculate

(a) the change in internal energy of the gas.

(b) the number of moles in the gas if the initial temperature is 300K.

(c) the molar heat capacity CP at constant pressure and

(d) the molar heat capacity CV at constant volume.


Answer:

Given:


(a) Pressure(P) = 2.0 × 105 Pa, dV(change in volume) = (200-100) cm3 = 10-4m3, since 1 m3 = 106 cm3


Heat(dQ) = 50 J.


Formula used:


Now we know, dQ = dU(change in internal energy) + dW(work)=


dU + PdV (first law of thermodynamics),


Where P = pressure, dV = change in volume.


=> dU = dQ - PdV = (50 - (2.0 × 105 x 10-4)) J = 30 J (Ans)


(b) For constant pressure, from equation of state PV/T = constant,


Where P = pressure, V = volume, T = temperature.


Hence we get: =, where


V1(initial volume) = 100 cm3, V2(final volume) = 200 cm3, T1 = 300K


=> .


Therefore, PdV = nRdT (for more than one mole),


Where P = pressure, dV = change in volume, n = number of moles, R = universal gas constant = 8.314 J/kg/mol, dT = change in temperature.


=> 2.0 × 105 x 10-4 = n x 8.314 x 300 (since T2-T1 = dT = 300K)


Therefore, n = 20/(R x 300) = 0.008 mol (Ans).


(c) dQ(heat) = 50 = nCpdT (at constant pressure),


Where n = number of moles, Cp = specific heat at constant pressure, dT = rise in temperature.


=> 50 = 0.008 x Cp x 300


=> Cp = 20.83 J/mol/K. (Ans)


(d) At constant volume, dU(change in internal energy) = dQ(heat) = nCvdT (since work done dW = PdV = 0, where P = pressure, dV = change in volume), from first law of thermodynamics.


n= number of moles, Cv = specific heat capacity at constant volume, dT = change in temperature.


=> 30 = 0.008 x Cv x 300


=> Cv = 12.5 J/mol/K. (Ans)



Question 8.

An amount Q of heat is added to a monatomic ideal gas in a process in which the gas performs a work Q/2 on its surrounding. Find the molar heat capacity for the process.


Answer:

Given: Amount of heat added(dQ) = Q


Amount of work done(dW) = Q/2.


Formula used:


dQ(heat) = dU(internal energy) + dW(work done).


Here, heat = Q and Work = Q/2(given)


=> .


We can write U = nCvdT and Q = nCdT, where n = no of moles, Cv = specific heat capacity at constant volume(when dQ = dU), C = molar heat capacity and dT = change in temperature.


Therefore, => C = 2Cv.


For a monoatomic ideal gas, we know that Cv = (3R/2) J/kg/mol,


Where R = universal gas constant = 8.314 J/kg/mol


Therefore, C = 2*(3R/2) = 3R J/kg/mol. (Ans)



Question 9.

An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation p = kV. Show that the molar heat capacity of the gas for the process is given by .


Answer:

Given:


P = kV …


Where,


P = pressure,


V = volume,


k = constant.


Formula used:


Equation of state of ideal gas:


PV = nRT = constant … (ii)


Where,


n is the number of moles of the gas,


R is the gas constant,


T is the temperature,


P = pressure, V2


T = temperature.


From (i), multiplying by dV on both sides:


PdV = kVdV.


Integrating from V = V1 to V2, we get



= with lower limit V1 and upper limit V2


=


Now, we know, PV = nRT - equation of state,


Where P = pressure, V = volume, n = number of moles, R = universal gas constant, T = temperature


Hence we can write, V1 = nRT1/P1. Since P1 = kV1, this becomes:


kV12 = nRT1. Similarly, KV22 = nRT2,


P1, V1, T1 - Pressure, volume, temperature of first gas


P2, V2, T2 - Pressure, volume, temperature of second gas


Therefore, subsituting, the above equation becomes:


= = … (iii)


Now, => (since P = kV) … (iv)


But Q = U + ഽPdV (first law of thermodynamics), where Q = heat, U = change in internal energy, W = total work done = ഽPdV


=> nCdT = nCvdT + (nR/2)dT


(since Q = nCdT and U = nCvdT)


=> C = Cv + nR/2 (proved),


Where


n = number of moles,


C = specific heat capacity,


Cv = specific heat capacity at constant volume,


R = universal gas constant,


dT = rise in temperature.



Question 10.

An ideal gas (CP/CV = γ) is taken through a process in which the pressure and the volume vary as p = αVb. Find the value of b for which the specific heat capacity in the process is zero.


Answer:

Given:


p = aVb


p = pressure, V = volume, a and b are constants.


Formula used:


We know, Q = U + ഽpdV from first law of thermodynamics,


where Q = change in heat, U = change in internal energy and ഽpdV = W = total work done, p = pressure, V = volume.


Since Q = nCdT, and U = nCvdT, we get


… (ii), n = no. of moles, C = specific


heat capacity, and Cv = specific heat capacity at constant volume,


dT = change in temperature,


Since specific heat capacity is 0(given),


…(iii)


(after integration of from V1 to V2)


Now, from equation of state, PV = nRT,


Where


P = pressure,


V = volume,


n = number of moles,


R = universal gas constant,


T = temperature.


Substituting p = aVb from (i):


aVb+1 = nRT


=> … (iv)


Substituting (iv) in (iii),



(Since (T2 - T1) = dT)


=>


=> b = -𝛾 (Ans)



Question 11.

Two ideal gases have the same value of CP/CV = γ. What will be the value of this ratio for a mixture of the two gases in the ratio 1: 2?


Answer:

We know Cp/Cv=γ, R=Cp-Cv,


where the molar heat capacity C, at constant pressure, is represented by Cp, at constant volume, the molar heat capacity C is represented by Cv and R is the universal gas constant.


Now,



For the first ideal gas,




Where Cp1 and CV1 is the molar heat capacity at constant pressure and constant volume


Similarly, for the second ideal gas




Where Cp2 and CV2 is the molar heat capacity at constant pressure and constant volume


Given,



i.e


dU1=nCV1dT


dU2=2nCV2Dt


When gas is mixed,





Also,



From (1) and (2)




Question 12.

A mixture contains 1 mole of helium (CP = 2.5 R, CV = 1.5R) and 1 mole of hydrogen (CP = 3.5 R, CV = 2.5 R). Calculate the values of CP, CV and γ for the mixture.


Answer:

Given,


CP1 = 2.5 R, CV1= 1.5R for helium


CP2 = 3.5 R, CV2= 2.5 R for hydrogen


n1=n2=1


We know dU =nCvdT


Where dU is the change in internal energy, n is the number of moles, Cv is the molar heat capacity at constant volume and dT is the change in temperature.


For the mixture,





Also,





Question 13.

Half mole of an ideal gas (γ =5/3) is taken through the cycle abcda as shown in figure. Take .



(a) Find the temperature of the gas in the states a, b, c and d.

(b) Find the amount of heat supplied in the processes ab and bc.

(c) Find the amount of heat liberated in the processes cd and da.


Answer:

Given, n=1/2, γ =5/3, R=25/3 J/Kmol


a) By ideal gas equation,


where P, V and T are the pressure, volume and absolute temperature; n is the number of moles of gas; R is the ideal gas constant.


Here temperature at a,


Here temperature at b,



Here temperature at c,



Here temperature at d,



b) Here ab is an isobaric process where heat supplied dQ can be expressed as



Here bc is an isochoric process where heat supplied dQ is



c) Heat liberated in cd, isobaric process dQ is



Heat liberated in da, isochoric process dQ is




Question 14.

An ideal gas (γ = 1.67) is taken through the process abc shown in figure. The temperature at the point a is 300K. Calculate

(a) the temperature at b and c,

(b) the work done in the process,

(c) the amount of heat supplied in the path ab and in the path bc and

(d) the change in the internal energy of the gas in the process.




Answer:

Let (P1, V1, T1), (P2, V2, T2), (P3, V3, T3) denote the pressure, volume and temperature at a, b and c respectively.


(a)For the process ab volume is constant


I.e. by ideal gas equation,





, temperature at b


For the process bc, pressure is constant.


By ideal gas equation,






temperature at c


(b)Here process ab is isochoric i.e. Wab=0


For process bc, P=200 kPa, change in volume is 50 cm3 from b to c



(c) From the first law of thermodynamics,



Where dQ is the amount of heat supplied


As ab is isochoric process dW=0






Here bc is an isobaric process where heat supplied dQ by first law of


Thermodynamics is







(d) dQ = dU + W


Now






Question 15.

In Joly’s differential steam calorimeter, 3g of an ideal gas is contained in a rigid closed sphere at 20°C. The sphere is heated by steam at 100°C and it is found that an extra 0.095 g of steam has condensed into water as the temperature of the gas becomes constant. Calculate the specific heat capacity of the gas in J g–1 K–1. The latent heat of vaporization of water = 540 cal g–1.


Answer:

Here,


m1 = Mass of gas present = 3 g, θ1 = 20°C, θ2 = 100°C


m2 = Mass of steam condensed = 0.095 g, L = 540 Cal/g = 540 × 4.2 J/g


In Joly’s differential steam calorimeter,





Question 16.

The volume of an ideal gas (γ = 1.5) is changed adiabatically from 4.00 litres to 3.00 litres. Find the ratio of

(a) the final pressure to the initial pressure and

(b) the final temperature to the initial temperature.


Answer:

Here γ= 1.5, V1=4, V2=3.Let P1 and P2 be the initial and final pressure


(a)Since it is an adiabatic process, So PVγ = const.


I.e.


(b)Also for an adiabatic process, TVγ -1=constant


I.e.






Question 17.

An ideal gas at pressure 2.5 × 105 Pa and temperature 300K occupies 100 cc. It is adiabatically compressed to half its original volume. Calculate

(a) the final pressure,

(b) the final temperature and

(c) the work done by the gas in the process. Take γ = 1.5.


Answer:

Here given, P1 = 2.5 × 105 Pa, V1 = 100 cc, T1 = 300 K, V2=50 cc


(a)Since it is an adiabatic process, So PVγ = const.


I.e.



(b) Also for an adiabatic process, TVγ -1=constant


I.e.





(c) Work done by the gas in the process






Question 18.

Air (γ = 1.4) is pumped at 2 atm pressure in a motor tyre at 20°C. If the tyre suddenly bursts, what would be the temperature of the air coming out of the tyre. Neglect any mixing with the atmospheric air.


Answer:

Given γ = 1.4,


Initial pressure, P1=2 atm


Final pressure, P2=1 atm


Initial Temperature, T1=20°C=293 K


Here bursting of tire is an adiabatic process,






Question 19.

A gas is enclosed in a cylindrical can fitted with a piston. The walls of the can and the piston are adiabatic. The initial pressure, volume and temperature of the gas are 100 kPa, 400 cm3 and 300 K respectively. The ratio of the specific heat capacities of the gas is CP/CV = 1.5. Find the pressure and the temperature of the gas if it is

(a) suddenly compressed

(b) slowly compressed to 100 cm3.


Answer:

Given, P1 = 100 KPa = 105 Pa, V1 = 400 cm3 , T1 = 300 K,


CP/CV = 1.5


(a)Suddenly compressed to V2 = 100 cm3 I.e. it is an adiabatic process


∴ PVγ = const.


I.e.



Also,





(b) Even though the container is slowly compressed the walls are adiabatic so heat transferred is 0.


Thus the values remain same


i.e. P2 = 800 KPa, T2 = 600 K.



Question 20.

The initial pressure and volume of a given mass of a gas (CP/CV = γ) are P0 and V0. The gas can exchange heat with the surrounding.

(a) It is slowly compressed to a volume V0/2 and then suddenly compressed to V0/4. Find the final pressure.

(b) If the gas is suddenly compressed form the volume V0 to V0/2 and then slowly compressed to V0/4, what will be the final pressure?


Answer:

Given CP/CV = γ


Let P1=P0 be the Initial Pressure, V1=V0 be the Initial Volume


(a)Since the volume is slowly compressed, temperature remains constant I.e. Isothermal compression. Let P2 and V2=V0/2 be the pressure and volume after slow compression





When volume is suddenly compressed, it is an adiabatic process .Let P3 and V3 be the pressure and volume after sudden compression .i.e.


PVγ = const.


I.e.


Substituting value of P2=2P0



(b) Since the volume is suddenly compressed, I.e. it is an adiabatic process Let P2 and V2=V0/2 be the pressure and volume after sudden compression, then


PVγ = const.


I.e.


Substituting value of P1=P0



Now, since the volume is slowly compressed, temperature remains constant I.e. Isothermal compression. Let P3 and V3=V0/4 be the pressure and volume after slow compression.






Question 21.

Consider a given sample of an ideal gas (CP/CV = γ) having initial pressure P0 and volume V0.

(a) The gas is isothermally taken to a pressure P0/2 and from there adiabatically to a pressure P0/4. Find the final volume.

(b) The gas is brought back to its initial state. It is adiabatically taken to a pressure P0/2 and from there isothermally to a pressure P0/4. Find the final volume.


Answer:

Given CP/CV = γ, initial pressure P1=P0 and initial volume V1= V0


a) Since the gas is isothermally taken to pressure P2=P0/2





Let P3=P0/4 and V3 be the pressure and volume after adiabatic compression.


Then,


PVγ = const.


I.e.



Substituting value of V2=2V0



(b)Here P1=P0, P2=P0/2 and the process is adiabatic. Let V1=V0 be the initial volume and V2 be the volume after process.


Then,


PVγ = const.


I.e.



I.e.


Let P3= P0/4 and V3 be the pressure and volume after isothermal process Then,






Question 22.

A sample of an ideal gas (γ = 1.5) is compressed adiabatically from a volume of 150 cm3 to 50 cm3. The initial pressure and the initial temperature are 150 kPa and 300 K. Find

(a) the number of moles of the gas in the sample,

(b) the molar heat capacity at constant volume,

(c) the final pressure and temperature,

(d) the work done by the gas in the process and

(e) the change in internal energy of the gas.


Answer:

Given P1 = 150 KPa = 150 × 103 Pa, V1 = 150 cm3 ,V2=50 cm3, T1 = 300 K


(a)By ideal gas equation,


where P, V and T are the pressure, volume and absolute temperature; n is the number of moles of gas; R is the ideal gas constant


I.e.



(b)We know Cp/Cv=γ R=Cp-Cv,


where the molar heat capacity C, at constant pressure, is represented by Cp, at constant volume, the molar heat capacity C is represented by Cv


Now,




(c) Since the process is adiabatic,


PVγ = const.


I.e.




Also, as the process is adiabatic,





(d) From the first law of thermodynamics,


, where dQ is the amount of heat supplied which is zero in an adiabatic process.


i.e.



, where n is the number of moles, Cv is the molar heat capacity at constant volume and dT is the change in temperature




(e)Change in internal energy, dU is



, where n is the number of moles, Cv is the molar heat capacity at constant volume and dT is the change in temperature.





Question 23.

Three samples A, B and C of the same gas (γ = 1.5) have equal volumes and temperatures. The volume of each sample is doubled, the process being isothermal for A, adiabatic for B and isobaric for C. If the final pressure is equal for the three samples, find the ratio of the initial pressures.


Answer:

Let VA, VB, VC be the volume of three gases and TA, TB, TCbe the temperature of A, B, C gas


Given, TA=TB=TC, VA=VB=VC


Here A is undergoing an isothermal process, where V1= VA, V2=2VA


Let P1 and P2 be the initial and final pressures,


Then,





Here B is adiabatic,


PVγ = const, where V1= VB, V2=2VB


Let P1 and P2 be the initial and final pressures,


I.e.




Here C is isobaric ,the pressure remains constant and equal to


Now, as the final pressures are equal for all the gases



, ratio of the initial pressures



Question 24.

Two samples A and B of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that γ satisfies the equation 1 – 21– γ = (γ – 1) ln2.


Answer:

Let P1 = Initial Pressure, V1 = Initial Volume, P2 = Final Pressure, V2 = Final Volume


Here A is expanded isothermally,


I.e. the work done,



Also, B is expanded adiabatically, i.e.



Given WA=WB


i.e.



In an adiabatic process,


PVγ = const,


I.e.



Substituting in (1)




We know, PV= nRT by ideal gas equation


i.e.



, the required relation



Question 25.

1 litre of an ideal gas (γ = 1.5) at 300 K is suddenly compressed to half its original volume.

(a) Find the ratio of the final pressure to the initial pressure.

(b) If the original pressure is 100 kPa, find the work done by the gas in the process.

(c) What is the change in internal energy?

(d) What is the final temperature?

(e) The gas is now cooled to 300 K keeping its pressure constant.

Calculate the work done during the process.

(f) The gas is now expanded isothermally to achieve its original volume of 1 litre. Calculate the work done by the gas.

(g) Calculate the total work done in the cycle.


Answer:

Given γ = 1.5 T=300 K, initial volume V1=1 L, Final volume V2=1/2 L. Let P1 and P2 be the initial and final pressures


(a)Here the process is adiabatic since volume is changed suddenly,


i.e.





(b)P1=100 KPa =105 Pa ,P2=21.5(105) KPa


Work done in adiabatic process,





(c) Here dQ=0, as it an adiabatic process


By first law of thermodynamics,



i.e.


(d)For an adiabatic process, let TI and T2 be initial and final temperature


TVγ -1=constant


I.e.






(e)Here the pressure is kept constant, i.e. isobaric


Work done in an isobaric process ,


Here,


Work done,


(f)Here the process is isothermal.


Work done,



(g)Work done in the cycle,



Question 26.

Figure shows a cylindrical tube with adiabatic walls and fitted with an adiabatic separator. The separator can be slid into the tube by an external mechanism. An ideal gas (γ= 1.5) is injected in the two aides at equal pressures and temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio 1 : 3. Find the ratio of the temperatures in the two parts of the vessel.




Answer:

Given:


The walls of the cylindrical tube and the separator are made with


adiabatic material. The separator can be slid in the tube by


external mechanism.


An ideal gas of is injected in the two aides of at equal pressure.


It is now slid to a position where it divides tube in the ratio 1:3.


The initial volume of the two aides are equal let’s say V/2,


Where, the total volume of the tube is V.


Now say the, left part of tube has V/4 volume and the right side has 3V/4 volume so that the ratio between them is 1:3.


In adiabatic process, (K = non zero constant)


Where P is the pressure of the gas and V is the volume and =


For ideal gas,


Where P is the pressure, V is the volume, T is the temperature of the gas and R is the gas constant and n is the number of moles of the gas.


Putting this in the adiabatic process condition we get,


(K’ is a non-zero constant)


Therefore,







Again for the other part of the tube,






As initially the gases were at the same pressure and volume, the temperatures would be the same as well.


Therefore,


Therefore, ,


Therefore the ratio of the final temperatures will be



Question 27.

Figure shows two rigid vessels A and B, each of volume 200 cm3 containing an ideal gas (CV = 12.5 J K–1 mol–1). The vessels are connected to a manometer tube containing mercury. The pressure in both the vessels is 75 cm of mercury and the temperature is 300 K.

(a) Find the number of moles of the gas in each vessel.

(b) 5.0 J of heat is supplied to the gas in the vessel A and 10 J to the gas in the vessel B. Assuming no appreciable transfer of heat from A to B calculate the difference in the heights of mercury in the two sides of the manometer. Gas constant R = 8.3 J K–1 mol–1.




Answer:

Given:


Gasses in both the vessels are at pressure of 75cm of mercury.


Therefore,


The volume of the vessel is


The temperature of the gas is 300K


(a) For Ideal gasses,


where P,V and T are the pressure, volume and temperature of the gas, n is the number of moles, and R is the gas constant.



Therefore, number of moles = 0.008


(b) The specific heat of the gas at constant volume is CV = 12.5 J K–1 mol–1


Therefore,


Where n is the number of moles, T is the rise in temperature, Q is the heat given.


Therefore, at constant volume, if we supply 5J and 10J heat to the vessels, the rise of temperature will be and


So the change in pressure in the vessels will be governed by



So for the first vessel, Change in pressure



For the second vessel


Therefore, the difference of pressure of the two vessels is



Which is equivalent to of mercury.


Therefore, the height of the mercury in the manometer tube is 12.5cm



Question 28.

Figure shows two vessels with adiabatic walls, one containing 0.1g of helium (γ = 1.67, M = 4 g mol–1) and the other containing some amount of hydrogen (γ= 1.4, M = 2g mol–1). Initially, the temperatures of the two gases are equal. The gases are electrically heated for some time during which equal amounts of heat are given to the two gases. It is found that the temperatures rise through the same amount in the two vessels. Calculate the mass of hydrogen.




Answer:

Given:


Two vessels with adiabatic walls, one contains 0.1g of helium (γ = 1.67, M = 4 g mol–1) and the other contains some amount of hydrogen (γ= 1.4, M = 2g mol–1)


The gasses are given the same amount of heat.


The temperature rises through the same amount.


0.1g of helium = 0.1/4 mole = 0.025mole


Let there be n moles of hydrogen in the other vessel.


= and so,


As the vessels are of constant volume there will be no work done by the gasses. The heat supplied will totally be used to increase internal energy.


Therefore, where Q is the heat supplied, n is the number of moles, Cv is the specific heat capacity of gas at constant volume, T is the change in temperature.


For helium,


For hydrogen, we assume for both cases the rise of temperature is T.


As per question,




⇒ n = 0.015


Again, Molar mass of hydrogen = 2g mol–1


Therefore, 0.015 mole of hydrogen hydrogen


Thus, there is 0.03g of hydrogen in the vessel.



Question 29.

Two vessels A and B of equal volume V0 are connected by a narrow tube which can be closed by a valve. The vessels are fitted with pistons which can be moved to change the volumes. Initially, the valve is open and the vessels contain an ideal gas (CP/CV = γ) at atmospheric pressure p0 and atmospheric temperature T0. The walls of the vessel A are diathermic and those of B are adiabatic. The valve is now closed and the pistons are slowly pulled out to increase the volumes of the vessels to double the original value.

(a) Find the temperatures and pressures in the two vessels.

(b) The valve is now opened for sufficient time so that the gases acquire a common temperature and pressure. Find the new values of the temperature and the pressure.


Answer:

Given:


Two vessels A and B of equal volume V0 are connected by a narrow tube which can be closed by a valve.


Vessels contain an ideal gas () at atmospheric pressure p0 and atmospheric temperature T0.


The walls of the vessel A are diathermic and those of B are adiabatic.


The pistons are slowly pulled out to increase the volumes of the vessels to double the original value.


(a) As the pistons are moved slowly to increase the volume, the expansion of gas in the diathermic vessel will be an isothermic process thus the temperature will be fixed at T0. P,V and T represent the pressure, volume and temperature of the gasses and subscripts 1 and 2 denote initial and final state respectively.


Thus,





For the adiabatic vessel,





Again for ideal gasses,




Thus the temperature and pressure in the diathermic vessel will T0 and P0/2 and in the adiabatic vessel, and .


(b) When the valve is open, the temperature will remain T0 throughout. Thus, there will be no change in temperature in the diathermic vessel so there will be change in pressure as well. For the gas in the diathermic vessel,


and for the adiabatic vessel


Therefore


Again,


Thus,


Thus the final temperature, when the valve is open will be T0 and the final pressure will be .



Question 30.

Figure shows an adiabatic cylindrical tube of volume V0 divided in two parts by a frictionless adiabatic separator. Initially, the separator is kept in the middle, an ideal gas at pressure p1 and temperature T1 is injected into the left part and another ideal gas at pressure p2 and temperature T2 is injected into the right part. CP/CV = γ is the same for both the gases. The separator is slid slowly and is released at a position where it can stay in equilibrium. Find

(a) the volumes of the two parts,

(b) the heat given to the gas in the left part

(c) the final common pressure of the gases.




Answer:

Given:


An adiabatic cylindrical tube of volume V0 is divided in two parts by a frictionless adiabatic separator.


An ideal gas at pressure p1 and temperature T1 is injected into the left part and another ideal gas at pressure p2 and temperature T2 is injected into the right part.


(a) When the piston is slowly moved to the equilibrium position, one side increases in volume when the other side decreases.


The processes will be adiabatic,


For the left part,


Where, subscript 1 and 2 represent the initial and the final state.


……….(1)


And for the right part,


……….(2)


We are assuming P to be the common pressure.


Dividing (1) by (2) we get,



Again,



So,




Therefore,


The final volume of the left and the right side will be and respectively.


(b) The heat given will be zero as the whole process is taking place in an adiabatic surrounding.


(c) So putting the above result in (1) we get,




Thus the final common pressure of the gasses will be



Question 31.

An adiabatic cylindrical tube of cross-sectional area 1 cm2 is closed at one end and fitted with a piston at the other end. The tube contains 0.03g of an ideal gas. At 1 atm pressure and at the temperature of the surrounding, the length of the gas column is 40 cm. The piston is suddenly pulled out to double the length of the column. The pressure f the gas falls to 0.355 atm. Find the speed of sound in the gas at atmospheric temperature.


Answer:

Given:


An adiabatic cylindrical tube of cross-sectional area 1 cm2 is closed at one end and fitted with a piston at the other end.


The tube contains 0.03g of an ideal gas at 1 atm pressure and at the temperature of the surrounding.


The length of the gas column is 40 cm.


The piston is suddenly pulled out to double the length of the column and the pressure of the gas falls to 0.355 atm.


The expansion process of the gas is adiabatic, so Where, subscript 1 and 2 represent the initial and the final state.


When the length is increased to double the volume is also doubled.


So,






The speed of sound in gas at atmospheric pressure is given as



where γ is the adiabatic constant, P is pressure and ρ is the volume density. The speed is






Question 32.

The speed of sound in hydrogen at 0°C is 1280 m s–1. The density of hydrogen at STP is 0.089 kg m–3. Calculate the molar heat capacities CP and CV of hydrogen.


Answer:

Given:


The speed of sound in hydrogen at 0°C is 1280 m s–1.


The density of hydrogen at STP is 0.089 kg m–3.


At STP the pressure P is 1.013×105Pa.


So the speed of sound in hydrogen where is the density of the gas.


So, putting data in, we get,


So, ;




Question 33.

4.0 g of helium occupies 22400 cm3 at STP. The specific heat capacity of helium at constant pressure is 5.0 cal K–1 mol–1. Calculate the speed of sound in helium at STP.


Answer:

Given:


4.0 g of helium occupies 22400 cm3 at STP.


The specific heat capacity of helium at constant pressure is 5.0 cal K–1 mol–1 = 21J K–1 mol–1





At STP the pressure P is 1.013×105Pa.


The velocity of sound will be where P is the pressure of the gas, V is the volume and M is the mass of the gas.


Thus putting the values given, we get,



The speed of sound in helium is m/s.



Question 34.

An ideal gas having density 1.7 × 10–3 g cm–3 at a pressure 1.5 × 105 Pa is filled in a Kundt tube. When the gas is resonated at a frequency of 3.0 kHz, nodes are formed at a separation of 6.0 cm. Calculate the molar heat capacities CP and CV of the gas.


Answer:

Given:


An ideal gas having density 1.7 × 10–3 g cm–3 = 1.7 kg m–3 at a pressure 1.5 × 105 Pa is filled in a Kundt tube.


When the gas is resonated at a frequency of 3.0 kHz, nodes are formed at a separation of 6.0 cm.


the node separation is given by which is 6.0cm. Therefore,



the frequency of the sound (f) is 3kHz


thus velocity of sound =


again,


where P is the pressure and is the density of the gas.


Thus





Question 35.

Standing waves of frequency 5.0 kHz are produced in a tube filled with oxygen at 300 K. The separation between the consecutive nodes is 3.3 cm. Calculate the specific heat capacities CP and CV of the gas




Answer:

Given:


Standing waves of frequency 5.0 kHz are produced in a tube filled with oxygen at 300 K.


The separation between the consecutive nodes is 3.3 cm.


the node separation is given by which is 3.3cm. Therefore,



the frequency of the sound is 5kHz


thus the velocity of sound will be


again,


where R is the gas constant, T is the temperature of the gas in Kelvin scale and M is the molar mass of the gas.


The molar mass of oxygen is


Thus putting in the values in the above expression,



Again,



Therefore, the CP and CV for oxygen are and respectively.