Laws Of Thermodynamics
Class 12th Concepts Of Physics Part 2 HC Verma Solution
Short Answer
Question 1.Should the internal energy of a system necessarily increase if heat is added to it?
Answer:
No, the internal energy of the system necessarily should not increase if heat is added to it.
Explanation
1. Internal energy is defined as the sum of the kinetic energy and potential energy of molecules of the system. It includes only the energy associated with the random motion of molecules of the system.
2. Random motion of molecules is associated with the temperature of the system. Thus, any change in temperature will change the internal energy of the system.
3. Change in internal energy is given as
ΔU=Cv ΔT
Where ΔU= change in internal energy
Cv=molar specific heat at constant volume
ΔT= change in temperature.
4. If ΔT =0 then, ΔU will also be zero.
5. In the isothermal process, where the change in temperature is zero, ΔU is also zero.
6. According to First law of thermodynamics,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
7. Since for an isothermal process ΔT=0, therefore ΔU=0. So, from the first law of thermodynamics ΔQ=ΔW.
8. For an isothermal process, heat supplied to the system is used up entirely in doing work.
Question 2.
Should the internal energy of a system necessarily increase if its temperature is increased?
Answer:
No, the internal energy of a system will not necessarily increase if its temperature is increased.
Explanation
1. Internal energy is defined as the sum of the kinetic energy and potential energy of molecules of the system. It includes only the energy associated with the random motion of molecules of the system.
2. The important thing about internal energy is that it depends only on the state of the system, not how that state was achieved.
3. Thus, the internal energy of a given mass of gas depends on its state described by specifics values of pressure, volume, and temperature.
4. The internal energy of ideal gas depends on only on temperature.
5. For system other than ideal gas, internal energy depends on pressure, volume and temperature combined.
Question 3.
A cylinder containing gas is lifted from the first floor to the second floor. What is the amount of work done on the gas? What is the amount of work done by the gas? Is the internal energy of the gas increased? Is the temperature of the gas increased?
Answer:
1. Work done on the gas will be zero.
Explanation
Atmospheric pressure decreases, as we increase the height.
So, when the cylinder is lifted from the first floor to second-floor atmospheric pressure on the cylinder will decrease. Hence, the volume of gas inside the cylinder will increase. In the expansion process, work is always done by the gas. So, in such case, no work will be done on the gas. But gas will do work.
2. Work done by the gas will be PΔV.
Explanation
When the cylinder is lifted from the first floor to second-floor atmospheric pressure on the cylinder will decrease. Hence the volume of gas inside the cylinder will increase. This means gas expands. In the expansion, process work is done by the gas.
We know that,
Work done = force ×displacement
Volume = area ×displacement
Therefore,
Work done=pressure ×volume
Let change in the volume of gas = ΔV
The pressure at which gas expands =P
Thus, work done by the gas W
W=PΔV
3. Change in internal energy and temperature will depend on the walls of the container, whether they are insulating or conducting. So, we cannot comment about internal energy and temperature unless we know the nature of the wall.
Question 4.
A force F is applied on a block of mass M. the block is displaced through a distance d in the direction of the force. What is the work done by the force on the block? Does the internal energy change because of this work?
Answer:
1. Work done by the force on the block is F
d.
Explanation
Given
Force =F
Displacement due to force = d
We know that
Work done= force × displacement × cos θ
θ in our case is zero as displacement is in the direction of the force.
Work done by the force = F ×d ×cos 0
=F ×d (∵ cos0=1)
∴ Work done by the force on the block is F ×d.
2. Internal energy will not change because of work done by the force on the block.
Explanation
Internal energy is defined as the sum of the kinetic energy and potential energy of molecules of the system. It includes only the energy associated with the random motion of molecules of the system.
If the block moves as a whole system with some velocity, the kinetic energy of the box is not to be included in internal energy.
Question 5.
The outer surface of a cylinder containing gas is rubbed vigorously by a polishing machine. The cylinder and its gas become warm. Is the energy transferred to the gas heat or work?
Answer:
Energy transferred to the gas is heat energy.
Explanation
Since the outer surface of the gas is rubbed and not displaced/shaken, no work will be done on the gas.
We know that,
Work done = force ×displacement
Force =pressure × area
∴work done = pressure × area × displacement
So, work will only be done when there is either displacement or the gas is expanded/compressed.
The cylinder and gas become warm because of the heat generated by friction between the polishing machine and surface of the cylinder. This heat will warm up the gas and outer surface of the cylinder.
Question 6.
When we rub our hands, they become warm. Have we supplied heat to the hands?
Answer:
Our hands become warm when we rub them due to the friction between our hands. Friction between our hands supplies heat to the hands.
Question 7.
A closed bottle contains some liquid. The bottle is shaken vigorously for 5 minutes. It is found that the temperature of the liquid is increased. Is heat transferred to the liquid? Is work done on the liquid? Neglect expansion on heating.
Answer:
Work is done when there is displacement in the body.
Work done = force ×displacement
As the bottle is shaken vigorously, that means the bottle is displaced from its position. And this displacement is due to external force. Therefore, work is done on the bottle. Due to this work, the temperature of the liquid increases. No external heat is supplied to the bottle and liquid.
Question 8.
The final volume of a system is equal to the initial volume in a certain process. Is the work done by the system necessarily zero? Is it necessarily nonzero?
Answer:
Work done by the system is neither necessarily zero nor nonzero when the final volume is equal to the initial volume.
Explanation
We know that,
Work done = force ×displacement
Volume = area ×displacement
Therefore,
Work done=pressure ×volume
Let change in the volume of system = ΔV = V2-V1
Pressure =P
Thus, work done by the system W at constant pressure
W=PΔV
But if V2=V1 then, ΔV=0
Work done by the system W=0
So, in an isobaric process (where pressure remains constant) work done by the system will be zero if initial and final volume are equal.
But we also know that in the cyclic process the system returns to its initial state. But still work done is not zero. In fact, in the cyclic process since the system returns to its initial state, internal energy becomes zero. This is because internal energy is a state variable. It depends on initial and final state only. And if initial and final state becomes equal, then change in internal energy will be zero.
According to First law of thermodynamics,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
In cyclic process ΔU=0.
So, from the first law of thermodynamics ΔQ=ΔW i.e. heat supplied to the system is converted entirely into work in a cyclic process.
Question 9.
Can work be done by a system without changing its volume?
Answer:
Yes, work can be done by a system without changing its volume if the process is cyclic.
Explanation
1. In cyclic process, the system returns to its initial state. Therefore, initial volume becomes equal to final volume and hence ΔV=0.
2. Also, since system returns to initial state, the internal energy of the system remains same.
3. This is because internal energy is a state variable. It depends on initial and final state only. And if initial and final state becomes equal, then change in internal energy will be zero.
4. So, for cyclic process ΔV=0 and ΔU=0.
5. According to First law of thermodynamics,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
6. So, from the first law of thermodynamics ΔQ=ΔW i.e. heat supplied to the system is converted entirely into work in a cyclic process.
7. Therefore, work can be done by a system without changing its volume.
Question 10.
An ideal gas is pumped into a rigid container having a diathermic wall so that the temperature remains constant. In a certain time interval, the pressure in the container is doubled. Is the internal energy of the contents of the container also doubled in the interval?
Answer:
Yes, the internal energy will also be doubled.
Explanation
1. Internal energy for an ideal gas is given as
U=nCvT
Where U= internal energy
N=number of moles
Cv=molar specific heat at constant volume
T= temperature
2. Since the ideal gas is continuously pumped into rigid container number of moles are also increasing.
3. When the pressure becomes doubled, the number of moles also gets doubled.
4. It is given that the temperature remains constant. So internal energy depends only on the number of moles.
5. Internal energy will also be doubled as the number of moles is getting doubled.
Question 11.
When a tyre bursts, the air coming out is cooler than the surrounding air. Explain.
Answer:
We know that pressure inside the tyre is greater than the atmospheric pressure of surroundings. So, when a tyre bursts there is an adiabatic expansion of the air (adiabatic because no heat is either supplied or released during expansion). In the expansion process, work is done by the gas. Now work done in an adiabatic process is given as
Where 
=number of moles
R=gas constant
T1=initial temperature
T2=final temperature
= ratio of specific heat at constant pressure and volume
Since work is done by the gas in adiabatic expansion, therefore, W>0. From equation (i), for W>0, T1>T2. This means that air
coming out of tyre will be at a lower temperature than when it was inside the tyre.
Question 12.
When we heat an object, it expands. Is work done by the object is this process? Is heat given to the object equal to the increase in its internal energy?
Answer:
1. We know that,
Work done = force ×displacement
Volume = area ×displacement
Therefore,
Work done=pressure ×volume
Let change in the volume of object = ΔV = V2-V1
Pressure =P
Thus, work done by the object W
W=PΔV
During expansion volume increases. Hence work done will be positive and equals PΔV.
2. According to First law of thermodynamics,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
Since work done by the gas is not zero, heat supplied to the object will not be equal to internal energy only.
Question 13.
When we stir a liquid vigorously, it becomes warm. Is it a reversible process?
Answer:
No, it is not a reversible process.
Explanation
1. When we stir a liquid vigorously, it becomes warm because we do work on the liquid and it increases its temperature.
2. To make it a reversible process, would require it to bring the temperature to its initial value by stirring the liquid in opposite direction. Only then we can call it a reversible process.
3. But we know that this is not possible. The only way we can decrease the temperature is by extracting heat from the liquid and not by stirring it in opposite direction.
Question 14.
What should be the condition for the efficiency of a Carnot engine to be equal to 1?
Answer:
We know that efficiency 
of a Carnot engine is
Where W=work done by the engine
Q1=heat absorbed by the engine
Q2=heat released by the engine
As seen from the formula (i), efficiency will be 1 when:
i) W=Q1
ii) Q2=0
Question 15.
When an object cools down, heat is withdrawn from it. Does the entropy of the object decrease in this process? If yes, is it a violation of the second law of thermodynamics state in terms of increase in entropy?
Answer:
1. The entropy of the system is the measure of molecular disorder or randomness, of a system.
2. When heat is withdrawn from the system, temperature decreases. This means that there is less randomness in the system. So, entropy will decrease.
3. But the heat withdrawn from the system is supplied to surroundings. Therefore, the entropy of the surroundings will increase.
4. The second law of thermodynamics states that net entropy of the universe always increases.
5. So, if the entropy of the system decreases, the entropy of the surroundings increases. Therefore, there is always a net increase in entropy.
6. Hence second law of thermodynamic is not violated in this case.
Objective I
Question 1.The first law of thermodynamics is a statement of
A. conservation of heat
B. conservation of work
C. conservation of momentum
D. conservation of energy
Answer:
According to First law of thermodynamics,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
It is a general law of conservation of energy applicable to any systemin which the energy transfer from or to surroundings is considered.
Question 2.
If heat is supplied to an ideal gas in an isothermal process,
A. the internal energy of the gas will increase
B. the gas will do positive work
C. the gas will do negative work
D. the said process is not possible.
Answer:
According to First law of thermodynamics,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
For an isothermal process ΔT=0, therefore ΔU=0.
Then,
ΔQ=ΔW
When heat is supplied to the system gas expands i.e. the volume of the gas increases.
We know that,
Work done = force ×displacement
Volume = area ×displacement
Therefore,
Work done=pressure ×volume
Let change in the volume of object = ΔV = V2-V1
Pressure =P
Thus, work done by the object W
W=PΔV
During expansion volume increases. Hence, work done will be positive and equals P
V.
Question 3.
Figure shows two process A and B on a system. Let ΔQ1 and ΔQ2 be the heat given to the system in processes A and B respectively. Then
A. ΔQ1 > ΔQ2
B. ΔQ1 = ΔQ2
C. ΔQ1 < ΔQ2
D. ΔQ1< ΔQ2
Answer:
Initial and final points of both processes A and B are same. Therefore, internal energy in both the processes will be the same because internal energy is a state variable, independent of the path taken.
The area under the P-V curve gives the work done on the system. From the graph, it can be seen the area under the curve for process A is more than the area under the curve for process B., therefore, work done on the system in process A ΔW1 is more than work done on the system in process B ΔW2
ΔW1 > ΔW2 …..(i)
According to First law of thermodynamics,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied/extracted to/from the system
ΔU=change in internal energy
ΔW=work done by/on the system
For process A
ΔQ1=ΔU+ΔW1 …(ii)
For process B
ΔQ2=ΔU+ΔW2 …(iii)
From equation (i) ,(ii) and (iii) it is clear that ΔQ1 > ΔQ2.
Question 4.
Refer to figure. Let ΔU1 and ΔU2 be the changes in internal energy of the system in the process A and B. Then
A. ΔU1 > ΔU2
B. ΔU1 = ΔU2
C. ΔU1 < ΔU2
D. ΔU1 ≠ ΔU2
Answer:
Initial and final points of both processes A and B are same. Therefore, change in internal energy in both the processes will be the same because internal energy is a state variable, independent of the path taken.
Therefore, ΔU1=ΔU2=0.
Question 5.
Consider the process on a system shown in the figure. During the process, the work done by the system.
A. continuously increases
B. continuously decreases
C. first increases then decreases
D. first decreases then increases.
Answer:
We know that,
Work done = force ×displacement
Volume = area ×displacement
Therefore,
Work done=pressure volume
Let change in the volume of system = ΔV = V2-V1
Pressure =P
Thus, work done by the system W at constant pressure
W=PΔV
From the graph we can see that V2 > V1 i.e. final volume is greater than the initial volume.
So, work done by the system continuously increases.
Question 6.
Consider the following two statements.
A. If heat is added to a system, its temperature must increase.
B. If positive work in done by a system in a thermodynamic process, its volume must increase.
A. Both A and B are correct
B. A is correct but B is wrong
C. B is correct, but A is wrong
D. Both A and B are wrong
Answer:
Statement A: For an isothermal process (where the temperature remains constant) if heat is added to a system temperature will not increases. So, statement A is wrong.
Statement B: We know that,
Work done = force ×displacement
Volume = area ×displacement
Therefore,
Work done=pressure ×volume
Let change in the volume of system = ΔV = V2-V1
Pressure =P
Thus, work done by the system W at constant pressure
W=PΔV
W>0 and positive only when V2 > V1.
Therefore, statement B is correct.
Question 7.
An ideal gas goes from the state i to the state f as shown in the figure. The work done by the gas during the process.
A. is positive
B. is negative
C. is zero
D. cannot be obtained from this information.
Answer:
Since the graph between P and T is a straight line passing through the origin, therefore P
T.
P can only be proportional to T when the volume is kept constant. This can be easily proved from the ideal gas equation which is
PV=RT
Since R is already a constant, if volume also becomes constant then PT.
Constant volume implies ΔV=0.
We know that,
Work done = force ×displacement
Volume = area ×displacement
Therefore,
Work done=pressure ×volume
Let change in the volume of system = ΔV = V2-V1
Pressure =P
Thus, work done by the system W
W=PΔV
For ΔV=0, W=0.
Question 8.
Consider two processes on a system as shown in the figure.
The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ΔW1 and ΔW2 be the work done by the system in the processes A and B respectively.
A. ΔW1 > ΔW2
B. ΔW1 = ΔW2
C. ΔW1 < ΔW2
D. Nothing can be said about the relation between ΔW1 and ΔW2.
Answer:
Given V1=V2
We know that,
Work done = force ×displacement
Volume = area ×displacement
Therefore,
Work done=pressure ×volume
Let change in the volume of system = ΔV = V2-V1
Pressure =P
Thus, work done by the system W
ΔW=PΔV
For process A ΔW1=P1ΔV1
For process B ΔW2=P2ΔV2
Since ΔV1=ΔV2 we can write,
P1 < P2 (from graph)
Therefore ΔW1 < ΔW2.
Question 9.
A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintained at this position. As time passes the pressure of the gas in the cylinder.
A. increases
B. decreases
C. remains constant
D. increases or decreases depending on the nature of the gas.
Answer:
1. When the piston is moved to compress the gas, the volume of the gas reduces and pressure and temperature increases.
2. After that piston is maintained at this position. So, volume becomes constant.
3. But during this sudden compression heat is generated. And since the cylinder is metallic this heat can escape out of the cylinder as metals are good conductors of heat.
4. When heat escapes out of the cylinder, the temperature of the gas will decrease which will, in turn, decrease the pressure of the gas.
Objective Ii
Question 1.The pressure p and volume V of an ideal gas both increase in a process.
A. Such a process is not possible.
B. The work done by the system is positive.
C. The temperature of the system must increase.
D. Heat supplied to the gas is equal to the change in internal energy.
Answer:
We know that,
Work done = force ×displacement
Volume = area ×displacement
Therefore,
Work done=pressure ×volume
Let change in the volume of system = ΔV = V2-V1
Pressure =P
Thus, work done by the system W
ΔW=PΔV
It is given that volume in increasing i.e. ΔV>0. So, ΔW>0 and positive.
From ideal gas equation, we know that
PV=nRT
Where P=pressure
V=volume
n=number of moles
R=gas constant
T=temperature
So, if both pressure and volume are increasing, then
the temperature must also increase as n and R are constant.
Question 2.
In a process on a system, the initial pressure and volume are equal to the final pressure and volume.
A. The initial temperature must be equal to the final temperature.
B. The initial internal energy must be equal to the final internal energy.
C. The net heat given to the system in the process must be zero.
D. The net work done by the system in the process must be zero
Answer:
From ideal gas equation, we know that
PV=nRT
Where P=pressure
V=volume
n=number of moles
R=gas constant
T=temperature
If PiVi=PfVf
Then nRTi=nRTf
∴ Ti=Tf
Since the initial state of the system (Pi, Vi, Ti) is equal to final state of the system (Pf, Vf, Tf), initial and final internal energy will also be equal as internal energy is a state variable.
Question 3.
A system can be taken from the initial state p1, V1 to the final state p2, V2 by two different methods. Let ΔQ and ΔW represent the heat given to the system and the work done by the system. Which of the following must be the same in both the methods?
A. ΔQ
B. ΔW
C. ΔQ + ΔW
D. ΔQ – ΔW
Answer:
According to First law of thermodynamics,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied/extracted to/from the system
ΔU=change in internal energy
ΔW=work done by/on the system
∴ ΔU=ΔQ – ΔW
It is given that initial state (p1, V1) and final state (p2, V2) is same for both the method.
So, change in internal energy will be the same for both the methods as internal energy is a state variable and is independent of the path taken to achieve a state.
Question 4.
Refer to figure. Let ΔU1 and ΔU2 be the change in internal energy in processes A and B respectively, ΔQ be the net heat given to the system in process A + B and ΔW be the net work done by the system in the process A + B.
A. ΔU1 + ΔU2 = 0
B. ΔU1 – ΔU2 = 0
C. ΔQ – ΔW = 0
D. ΔQ + ΔW = 0
Answer:
As seen from the figure initial state of A is the final state of B and vice-versa.
So, the change in internal energy of process A and B will be equal in magnitude but with a negative sign.
So ΔU1=-ΔU2
∴ ΔU1+ΔU2=0.
Question 5.
The internal energy of an ideal gas decreases by the same amount as the work done by the system.
A. The process must be adiabatic
B. The process must be isothermal
C. The process must be isobaric
D. The temperature must decrease
Answer:
1. The internal energy of an ideal gas is proportional to the temperature. Since internal energy is decreasing, therefore, the temperature must decrease.
2. According to the question, change in internal energy is equal to change in work done.
3. According to First law of thermodynamics,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied/extracted to/from the system
ΔU=change in internal energy
ΔW=work done by/on the system
4. We know for adiabatic process ΔQ=0. So, from first law ΔU=ΔW.
Exercises
Question 1.A thermally insulated, closed copper vessel contains water at 15°C. When the vessel is shaken vigorously for 15 minutes, the temperature rises to 17°C. The mass of the vessel is 100 g and that of the water is 200 g. The specific heat capacities of copper and water are 420 J kg–1 K–1 and 4200 J kg–1 K–1 respectively.
Neglect any thermal expansion.
(a) How much heat is transferred to the liquid-vessel system?
(b) How much work has been done on this system?
(c) How much is the increase in internal energy of the system?
Answer:
Given
Initial temperature of water T1=15℃ =288K
Final temperature of water T2=17℃ =290K
Specific heat capacity of copper cc=420 J kg–1 K–1
Specific heat capacity of water cw=4200 J kg–1 K–1
Mass of copper vessel mc=100g = 100×10-3kg
Mass of water mw=200g = 200×10-3kg
a) It is given that copper vessel is thermally insulated. Therefore, no heat from the surroundings can be transferred to the liquid-vessel system.
b) Work done on this system will be
ΔW=mwcwΔT + mcccΔT
ΔT=T2-T1=290-288=2K
So,
ΔW= 200×10-3×4200×2 + 100×10-3×420×2
ΔW=1680 + 84=1764J
∴ work done on the system is 1764J.
c) From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
From part (a) we have concluded that ΔQ=0
So, first law becomes
ΔU=-ΔW
Work done on the system ΔW=1764 J
So, work done by the system ΔW=-1764J
ΔU = -(-1764) = 1764J
∴ Increase in internal energy of the system = 1764J.
Question 2.
Figure shows a paddle wheel coupled to a mass of 12 kg through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity 4200 JK–1 kept in an adiabatic container. Consider a time interval in which the 12 kg block falls slowly through 70 cm.
(a) How much heat is given to the liquid?
(b) How much work is done on the liquid?
(c) Calculate the rise in the temperature of the liquid neglecting the heat capacity of the container and the paddle.
Answer:
Given
Mass attaches to the pulley m = 12kg
Heat capacity of liquid s =4200 JK-1
Height through which mass fall= 70cm=0.7m
a) Paddle immersed in liquid is kept in an adiabatic container. So, no heat can be either supplied or extracted to the liquid. Therefore, heat given to liquid is zero.
b) As no heat is supplied to liquid and pulley is frictionless, work done on the liquid will be equal to the potential energy of the mass attached to the pulley.
Work done on the liquid = potential energy of mass
Potential energy of mass= mgh
Where g=acceleration due to gravity=10ms-2
Potential energy of mass= 12×10×0.7=84J
∴ work done on liquid =84J.
c) The mechanical work calculated in the second part will be converted into heat. This heat will be supplied to liquid and due to which temperature of the liquid will rise.
We know that,
Heat capacity 
Where ΔQ = heat supplied
ΔT=rise in temperature
Since work done is equal to heat supplied, therefore
∴ the rise in temperature of the liquid will be 0.02K.
Question 3.
A 100 kg block is started with a speed of 2.0 m
s–1 on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20.
(a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt.
(b) Consider the situation com a frame of reference moving at 2.0 m s–1 along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 ms–1. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt.
(c) Find the work done in this frame by the external force holding the belt.
Answer:
Given
Mass of block m=100kg
Initial velocity u=2 m/s
Coefficient of kinetic friction μ = 0.20
a) Since the block comes to stop final velocity will be zero.
So, final velocity v=0 m/s
When the block is moving over belt there is kinetic friction between the lower surface of the block and upper surface of the belt. And we know that heat is produced due to friction between two surfaces. Now because of this heat, the internal energy of block will change.
So,
Kinetic energy lost in heat due to friction = change in the internal energy
Kinetic energy lost = initial kinetic energy- final kinetic energy
change in internal energy is 200J.
b) Given
The velocity of the frame of reference uo= 2m/s
So, in this frame of reference initial and final velocity of the block will change.
New initial velocity u’=u-uo =2-2 = 0m/s
New final velocity v’ = v-uo =0-2 =-2m/s
So, increase in kinetic energy = (final – initial) kinetic energy
Increase in kinetic energy in com frame of reference is 200J.
c) Total work done in com frame of reference will be work done due to friction plus work done to give final velocity.
We know that force of friction f=N
Where = coefficient of friction =0.02
N=normal reaction =mg
f = 0.02×100×10 = 200N
from newton’s second law of motion
force= mass ×acceleration
so,
200=100×acceleration
Using the third equation of motion,
v’2-u’2=2as
where s=displacement as seen in com frame of reference
(-2)2-0=2×2×s
s=1m
work done due to friction Wf =force ×displacement
Wf=200×1=200J
Now to calculate work done to give final velocity, we will work-energy theorem.
According to work-energy theorem,
Work done = change in kinetic energy
So,
Total work done W=W’ + Wf =200+200=400J.
∴ work done in com frame of reference= 400J.
Question 4.
Calculate the change in internal energy of a gas kept in a rigid container when 100 J of heat is supplied to it.
Answer:
Given
Heat supplied ΔQ = 100J
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
We know that,
Work done = force ×displacement
Volume = areadisplacement
Therefore,
Work done=pressure ×volume
Let change in the volume of system = ΔV = V2-V1
Pressure =P
Thus, work done by the gas
ΔW=PΔV
Since the gas is kept in a rigid container, therefore ΔV=0 in this case.
So, ΔW=0
Thus, first law will become
ΔQ=ΔU = 100J
∴ change in internal energy will be 100J.
Question 5.
The pressure of gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc.
(a) Calculate the work done by the gas.
(b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas?
Answer:
Given
Initial pressure P1=10kPa=10×103Pa
Final pressure P2=50kPa=50×103Pa
Initial volume V1=200cc=200×10-6m3
Final volume V2=50cc= 50×10-6m3
a) We know that,
Work done = force ×displacement
Volume = area ×displacement
Therefore,
Work done=pressure ×volume
Let change in the volume of system = ΔV = V2-V1
Pressure =P
Thus, work done by the gas
ΔW=PΔV
Here we have given two values of pressure. So, we will take the average value of pressure
Average pressure P
Therefore, ΔW = 30×103× (50-200)×10-6
ΔW= -4.5J
Work done by the gas is -4.5J.
b) Given that no heat is supplied or extracted from the gas.
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
Since ΔQ=0
Therefore ΔU = -ΔW = -(-4.5) J=4.5J
∴ the change in internal energy of the gas is 4.5J.
Question 6.
An ideal gas is taken from an initial state i to a final state f in such a way that the ratio of the pressure to the absolute temperature remains constant. What will be the work done by gas?
Answer:
Given
Initial pressure Pi
Final pressure Pf
Initial temperature Ti
Final temperature Tf
It is given that
From ideal gas equation, we know
PV=nRT
Where V=volume of gas
R=gas constant
n= number of moles
applying ideal gas equation for both processes, we get
And
From equation (i), (ii) and (iii), we get
∴ Vi=Vf
We know that,
Work done = force ×displacement
Volume = area ×displacement
Therefore,
Work done=pressure × volume
Let change in the volume of system = ΔV = Vf-Vi
Pressure =P
Thus, work done by the gas
ΔW=PΔV=P (Vf-Vi) =0
∴ work done by the gas is zero.
Question 7.
Figure shows three paths through which a gas can be taken from the state A to the state B. Calculate the work done by the gas in each of the three paths.
Answer:
We know that work done by the gas is given as
ΔW=PΔV
From graph we can write
VA =VC = 10cc=10×10-6m3
VD=VB = 25cc= 25×10-6m3
PB=PC=30kPa= 30×103Pa
PA=PD=10kPa=10×103Pa
Work done in path ADB WADB=WAD + WDB
WADB = PA (VD-VA) + 0 (∵ WDB =0 because VD=VB)
=10×103×(25-10)×10-6
=0.15J
Work done in path AB WAB= Pavg(VB-VA)
WAB=20×103× (25-10)×10-6
=0.30J
Work done in path ACB WACB= WAC+WBC
WACB = 0+PB(VB-VC) (∵ WAC=0 because VA=VC)
=30×103×(25-10)×10-6
=0.45J
Question 8.
When a system is taken through the process abc shown in figure 80J of heat is absorbed by the system and 30 J of work is done by it. If the system does 10 J of work during the process adc, how much heat flows into it during the process?
Answer:
Given
Heat absorbed in process abc ΔQ1=80J
Work done by the system in process abc ΔW1=30J
Work done by the system in process adc ΔW2=10J
Let heat absorbed into the system during process adc =ΔQ2
Now initial point a and final point c is the same for both the processes is the same. So, change in internal energy will be the same for both the process, as internal energy is a state function independent of the path taken.
Therefore,
ΔU1=ΔU2=ΔU ……(i)
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
Using first law of thermodynamics for process abc
ΔQ1=ΔU1+ΔW1
ΔU1=ΔQ1-ΔW1=80-30=50J
Using first law of thermodynamics for process adc
ΔQ2=ΔU2+ΔW2
=ΔU1+ΔW2 (from (i))
= 50+10=60J
∴ heat absorbed into the system during process adc= 60J.
Question 9.
50 cal of heat should be supplied to take a system from the state A to the state B through the path ACB as shown in the figure. Find the quantity of heat to be supplied to take it from A to B via ADB.
Answer:
From graph we can write
VA =VD = 200cc=200×10-6m3
VB=VC = 400cc= 400×10-6m3
PB=PD=155kPa= 155×103Pa
PA=PC=50kPa=50×103Pa
Given
Heat absorbed in process ABC ΔQ1=50cal= 50×4.2 J=210J
Let heat absorbed into the system during process ADC =ΔQ2
We know that work done by the gas is given as
ΔW=PΔV
Work done in path ACB WACB =ΔW1= WAC+WBC
ΔW1=PA(VC-VA) + 0 (WBC=0 because VB=VC)
=50×103×(400-200)×10-6
=10J
Work done in path ADB WADB=ΔW2=WAD + WDB
ΔW2 = PB (VD-VB) +0 (∵ WAD=0 because VA=VD)
=155×103×(400-200)×10-6
=31 J
Now initial point A and final point C is the same for both the processes is the same. So, change in internal energy will be the same for both the process, as internal energy is a state function independent of the path taken.
Therefore,
ΔU1=ΔU2=ΔU ……(i)
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
Using first law of thermodynamics for process ABC
ΔQ1=ΔU1+ΔW1
ΔU1=ΔQ1-ΔW1=210-10=200J
Using first law of thermodynamics for process adc
ΔQ2=ΔU2+ΔW2
=ΔU1+ΔW2 (from (i))
=200+31=231J.
∴ heat supplied to the system during process ADC=231J.
Question 10.
Calculate the heat absorbed by a system in going through the cyclic process shown in the figure.
Answer:
We know that in the cyclic process the system returns to its initial state. So, change internal energy in the cyclic process will be zero as internal energy is a state function.
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
Since ΔU=0, first law becomes
ΔQ=ΔW
In a PV graph work done is equal to the area under the curve.
ΔW= area of a circle
ΔQ=ΔW=area of circle
Diameter of the circle =300-100=200
Area of the circle =π × (radius)2
=π × 100×100×10-6×103
(10-6×103 is because volume and pressure are given in cc and kPa respectively)
Area of circle =3.14×10=31.4
∴ Heat absorbed by a system = 31.4J.
Question 11.
A gas is taken through a cyclic process ABCA as shown in the figure. If 2.4 cal of heat is given in the process, what is the value of J?
Answer:
‘J’ is mechanical equivalent of heat a conversion factor between two different units of energy: calorie to joule
From the graph we can write
VA=VB=500cc=500×10-6m3
VC=700cc=700×10-6m3
PA=PC=100kPa=100×103 Pa
PB=200kPa=200×103Pa
We know that work done by the gas is given as
ΔW=PΔV
Work done in path AB=0 as VA=VB.
Work done in path CA ΔW1=PA(VA-VC)
=100×103× (500-700)×10-6
=-20J
Work done in path BC ΔW2= Pavg(VC-VB)
ΔW2=150×103×(700-500)×10-6
=30J
Total work done in process ABCA=ΔW=ΔW1+ΔW2
=30-20=10J
We know that in the cyclic process the system returns to its initial state. So, change internal energy in the cyclic process will be zero as internal energy is a state function.
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
Since ΔU=0, first law becomes
ΔQ=ΔW=10J
But it is given in question that ΔQ=2.4cal
So, 2.4×J=10Joule
∴ value of ‘J’ is 4.17J/cal.
Question 12.
A substance is taken through the process abc as shown in the figure. If the internal energy of the substance increases by 5000 J and heat of 2625 cal is given to the system, calculate the value of J.
Answer:
‘J' is mechanical equivalent of heat a conversion factor between two different units of energy: calorie to the joule.
Given
Heat given to system =2625cal =2625×J J
Change in internal energy= 5000J
From graph
Va=0.02m3
Vb=Vc=0.05m3
Pa=Pb=200kPa=200×103Pa
Pc=300kPa=300×103Pa
We know that work done by the gas is given as
ΔW=PΔV
Work done in process abc=ΔW=Wab+Wbc
ΔW=Pa(Vb-Va)+0 (∵ Wbc=0 because Vb=Vc)
ΔW=200×103×(0.05-0.02)=6000J
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
ΔQ=5000+6000=11000J
But ΔQ=2625cal =2625×J J
Therefore,
2625×J=11000
∴ value of ‘J’ is 4.19joule/cal.
Question 13.
A gas is taken along the path AB as shown in the figure. If 70 cal of heat is extracted from the gas in the process, calculate the change in the internal energy of the system.
Answer:
Given
Heat extracted from the system ΔQ=-70cal=-70×4.2=-294J
The negative sign is because heat is extracted from the system.
From graph
VA=250cc=250×10-6m3
VB=100cc=100×10-6m3
PA=200kPa=200×103Pa
PB=500kPa=500×103Pa
We know that work done by the gas is given as
ΔW=PΔV
Here since we have two values of pressure we will take average pressure.
ΔW=Pavg(VA-VB)
= 350×103× (250-100) ×10-6
=-52.5J
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
Therefore,
ΔU=-294-(-52.5) =-241.5J
∴ change in internal energy is -241.5J.
Question 14.
The internal energy of a gas is given by U = 1.5 pV. It expands from 100 cm3 to 200 cm3 against a constant pressure of 1.0 × 105 Pa. Calculate the heat absorbed by the gas in the process.
Answer:
Given
Constant pressure p=1.0×105 Pa
Change in volume ΔV=(200-100)×10-6m3=10-4m3
Internal energy U= 1.5pV
So, change in internal energy ΔU=1.5pΔV
=1.5×1.0×105×10-4
=15J
We know that work done by the gas is given as
ΔW=pΔV=1.0×105×10-4=10J
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
Therefore,
ΔQ=15+10=25J
∴ heat absorbed by the system=25J.
Question 15.
A gas is enclosed in a cylindrical vessel fitted with a frictionless piston. The gas is slowly heated for some time. During the process, 10 J of heat is supplied, and the piston is found to move out 10 cm. Find the increase in the internal energy of the gas. The area of cross section of the cylinder = 4 cm2 and the atmospheric pressure = 100 kPa.
Answer:
Given
Heat supplied to system ΔQ = 10J
Atmospheric pressure P= 100kPa =100×103Pa
Displacement of the piston d=10cm=0.1m
Area of cross section of cylinder A=4cm2 =410-4m2
Gas will expand when is heat to the system. Therefore, the volume of gas expanded ΔV
ΔV=area ×displacement
=A ×d
ΔV=4×0.1×10-4=40×10-6m3
We know that work done by the gas is given as
ΔW=PΔV
=100×103×40×10-6
ΔW =4J
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
ΔU=ΔQ-ΔW = 10-4 =6J
Thus, the increase in internal energy is 6J.
Question 16.
A gas is initially at a pressure of 100 kPa and its volume is 2.0 m3. Its pressure is kept constant and the volume is changed from 2.0 m3 to 2.5 m3. Its volume is now kept constant and the pressure is increased from 100 kPa to 200 kPa. The gas is brought back to its initial state, the pressure varying linearly with its volume.
(a) Whether the heat is supplied to or extracted from the gas in the complete cycle?
(b) How much heat was supplied or extracted?
Answer:
From graph
Va=2m3
Vb=Vc=2.5m3
Pa=Pb=100kPa=100103Pa
Pc=200kPa=200103Pa
Work done in process ABCA = area enclosed by the triangle ABC
ΔW =0.5×BC×AB
Where BC= height of the triangle
AB= base of triangle
So,
ΔW=0.5×(200-100)×103(2.5-2)
ΔW=0.5×100×0.5×103 =25000J
a) Process ABCA is a cyclic process. The system is brought back to its initial state. Since internal energy is a state function, change in internal energy will be zero.
So, ΔU=0
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
∴ ΔQ=ΔW=25000J
Since work done is positive, work is done by the gas. When work is done by the gas heat is supplied to the system.
b) Amount of heat supplied = work done by the gas in the cyclic process
∴ ΔQ=ΔW=25000J.
Question 17.
Consider the cyclic process ABCA, shown in the figure, performed on a sample 2.0 mol of an ideal gas. A total of 1200 J of heat is withdrawn from the sample in the process. Find the work done by the gas during the part BC.
Answer:
Given
Heat extracted from the system ΔQ=-1200J
(negative sign is because heat is extracted from the system)
Number of moles of the gas n=2.0
From the graph we can write
TA=300K
TB=500K
VA=VC
We know that work done by the gas is given as
ΔW=PΔV
Where ΔV =change in volume
P =pressure
So, work done along line CA will be zero, as VA=VC.
Thus, total work done will be
ΔW=WAB+WBC
WAB = P(VB-VA)
But we know that ideal gas equation is
PV=nRT
Where n= number of moles
R=gas constant
T=temperature
∴ PΔV=nRΔT
Thus, we can write WAB = P(VB-VA) = nR(TB-TA)
∴ ΔW= nR(TB-TA) + WBC
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
Process ABCA is a cyclic process. The system is brought back to its initial state. Since internal energy is a state function, change in internal energy will be zero.
So, ΔU=0.
So, first law becomes
ΔQ=ΔW
ΔQ= nR(TB-TA) + WBC
We know that R=8.31J/K mol
WBC =ΔQ- nR(TB-TA)
= -1200 - 2×8.31× (500-300)
=-1200 – 3324
=-4524J
Thus, work done along path BC =-4524J
Question 18.
Figure shows the variation in the internal energy U with the volume V of 2.0 mol of an ideal gas in a cyclic process abcda. The temperatures of the gas at b and c are 500 K and 300 K respectively. Calculate the heat absorbed by the gas during the process.
Answer:
Given
Number of moles n=2
Temperature at b Tb=500K
Temperature at c Tc=300K
From the graph it is clear that
Tb=Ta and Td=Tc
Thus, path ab and cd are isothermal paths.
We know that work done by the gas is given as
ΔW=PΔV
Where ΔV =change in volume
P =pressure
Again, from the graph we can see that ΔV=0 for path bc and da.
Therefore, work done along path bc and da are zero.
So, total work done ΔW=Wab+Wcd
We know that work done in an isothermal process is given as
Where n=number of moles
R=gas constant =8.31J/Kmol
T=temperature
Vf=final volume
Vi=initial volume
Wab = 8310×ln2
Similarly,
Wcd = -4986 ×ln2
So, total work done ΔW=Wab+Wcd
=8310×ln2 - 4986×ln2
=3324×ln2
ΔW =3324×0.693=2304.02J
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
Process ABCA is a cyclic process. The system is brought back to its initial state. Since internal energy is a state function, change in internal energy will be zero.
So, ΔU=0.
So, first law becomes
ΔQ=ΔW=2304.02J
Thus, heat absorbed by the system is 2304.02J.
Question 19.
Find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J kg–1 K–1 and its densities at 0°C and 4°C are 999.9 kg m–3 and 1000 kg m–3 respectively. Atmospheric pressure = 105 Pa.
Answer:
Given
Mass of water m = 2kg
Change in temperature T =4℃ -0℃ =4℃
Specific heat capacity of water c =4200 J kg–1 K–1
Density of water at 0°C=999.9 kg m–3
Density of water at 4°C=1000 kg m–3
Atmospheric pressure P = 105 Pa.
We know that specific heat capacity is given by
Where ΔQ = heat supplied to the system
Therefore, ΔQ=cmΔT
=4200×2×4=33600J
We know that work done by the gas is given as
ΔW=PΔV
Where ΔV =change in volume
P =pressure
Also,
Volume at 0oC V1
Similarly, volume at 4oC V2
∴ ΔW=P(V2-V1)
From first law of thermodynamics, we know that,
Q=U+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
ΔU=ΔQ-ΔW
=33600-(-0.02)
=33599.98J
Thus, change in internal energy is 33599.98J.
Question 20.
Calculate the increase in the internal energy of 10g of water when it is heated from 0°C to 100°C and converted into steam at 100 kPa. The density of steam = 0.6 kg m–3. The specific heat capacity of water = 4200 J kg–1 °C–1 and the latent heat of vaporization of water = 2.25 × 106 J kg–1.
Answer:
Given
The density of steam ρ’= 0.6 kg m–3
Mass of water m=10g =0.010kg
Specific heat capacity of water c = 4200 J kg–1 °C–1
latent heat of vaporization of water L = 2.25 × 106 J kg–1.
Pressure P =100kPa =100×105Pa
Change in temperature ΔT= (100-0) oC =100oC
Density of water ρ =1000 kg m-3
We know that specific heat capacity is given by
Where ΔQ = heat supplied to the system
Therefore, ΔQ= cmΔT
Also, ΔQ =mL
Where m= mass of the substance
L=latent heat
Therefore, ΔQ=mL + cmΔT
=0.010×2.25 × 106 + 4200×0.01×100
=22500+4200
=26700J
We know that work done by the gas is given as
ΔW=PΔV
Where ΔV =change in volume
P =pressure
Also,
ΔW=105×0.01699=1699 J
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
ΔU=ΔQ-ΔW
=26700-1699
=25001J
Thus, change in internal energy is 25001J.
Question 21.
Figure shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5 nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p1, T1 on the left and p2, T2 on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides.
(a) How much work has been done by the gas on the left part?
(b) Find the final pressures on the two sides.
(c) Find the final equilibrium temperature.
(d) How much heat has flown from the gas on the right to the gas on the left?
Answer:
a) According to the question, the diathermic separator between both the part is fixed. So, no change in volume will be observed. And as we know work done on gas is PV. Therefore, no work will be done on the left part during the process as the volume is not changing.
First, we will calculate the final temperature and then final pressure.
(c) Given
Pressure of left chamber =P1
Pressure of right chamber =P2
Temperature of left chamber =T1
Temperature of right chamber =T2
Let the number of moles in the left chamber be n1
Number of moles in the right chamber be n2
Diathermic wall has divided the tube in two equal part. So, the volume of the left and the right chamber will be V/2.
Applying ideal gas equation in the left chamber,
Similarly applying ideal gas equation in the right chamber,
Total number of moles n=n1 +n2
The internal energy of ideal gas is given as
U=nCvT
Where Cv=molar specific heat at constant volume
T=temperature.
According to question,
U=1.5nRT
nCvT=1.5nRT
So, Cv=1.5R
The internal energy of the left chamber U1=n1CvT1
The internal energy of right chamber U2= n2CvT2
Total internal energy U= U1+U2
1.5nRT= n1CvT1+ n2CvT2
1.5nRT=Cv(n1T1+n2T2)
Substituting the value of Cv
1.5nRT=1.5R(n1T1+n2T2)
nT=n1T1+n2T2
substituting the value of n1 and n2 in above equation
Substituting the value of n in the above equation,
Thus, final equilibrium temperature T=
.
(b) Now we will find final pressures on both sides.
Let final pressure of left chamber P1’
Final pressure of right chamber P2’
Applying ideal gas equation in the left chamber before and after equilibrium
From equation (i) and (ii),
Substituting the value of T,
Applying ideal gas equation in the right chamber before and after equilibrium
From equation (iii) and (iv),
Substituting the value of T,
(d) The internal energy of ideal gas is given as
U=nCvT
Where Cv=molar specific heat at constant volume
T=temperature.
As stated in part (a) no work will be done on either chamber of the vessel as the diathermic separator is fixed.
So, ΔW=0 for the right chamber of the tube.
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
ΔQ=ΔU
Change in internal energy of the right chamber after equilibrium has reached will be
ΔU=n2CvT2 -n2CvT
Substituting the value of n2, Cv and T in the above equation
Thus, heat flown from left to right chamber is
.
Question 22.
An adiabatic vessel of total volume V is divided into two equal parts by a conducting separator. The separator is fixed in this position. The part on the left contains one mole of an ideal gas (U = 1.5 nRT) and the part on the right contains two moles of the same gas. Initially, the pressure on each side is p. The system is left for sufficient time so that a steady state is reached. Find
(a) the work done by the gas in the left part during the process.
(b) the temperature on the two sides in the beginning,
(c) the final common temperature reached by the gases,
(d) the heat given to the gas in the right part and
(e) the increase in the internal energy of the gas in the left part.
Answer:
a) According to the question, conducting separator between both the part is fixed. So, no change in volume will be observed. And as we know work done a gas is PV. Therefore, no work will be done on the left part during the process as the volume is not changing.
b) Given
The vessel is divided into two equal parts.
So, the volume of left part V1 and volume of right part V2 will half of the total volume.
Initial pressure on each side is p
Number of moles on left side n1 =1
Number of moles on right side n2=2
Let initial temperature for the left and right part be T1 and T2 respectively.
For the left part, applying the ideal gas equation
Similarly, for the right part
c) It is given that the internal energy of the gas is
U=1.5nRT
Where n=number of moles
R=gas constant
T=final equilibrium temperature of both parts
The internal energy of ideal gas is given as
U=nCvT
Where Cv=molar specific heat at constant volume
T=temperature.
Total moles n=n1+n2=1+2=3
U=3CvT
Let U1 and U2 be the internal energy of the left and right part respectively.
So, U1=n1CvT1= CvT1
And U2=n2CvT2=2CvT2
(gas is same on both the part so Cv will be same)
Now
U=U1+U2
3CvT = CvT1+ 2CvT2
3T=T1+2T2
d) As stated in part (a) no work will be done on either part of the vessel as conducting separator is fixed.
So, ΔW=0 for the right part of the vessel.
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
∴ ΔQ=ΔU
It is given that the internal energy of the gas is
U=1.5nRT
Where n=number of moles
R=gas constant
T=final equilibrium temperature of both parts
ΔU=1.5nRΔT
For the right part, change in internal energy after equilibrium has reached will be due to the change in temperature from T2 to T.
ΔU=1.5n2R(T-T2)
Thus, heat given to the right part is 
.
e) Since ΔW is zero as the volume is fixed, therefore, the first law of thermodynamics.
But since heat is given to the right part that means heat is extracted from the left part. So, the internal energy of the left part of will decrease.
Therefore, heat given to the right part will be equal to the negative internal energy of the left part.
ΔQ=-ΔU (left part)
.