H C VERMA Solutions for Class 12-science Physics Chapter 4 - Laws of Thermodynamics

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Chapter 4 - Laws of Thermodynamics Exercise 62

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 100g and that of the water is 200g. The specific heat capacities of copper and water are 420J/kg-K and 4200J/kg-K 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?

Solution 1

(a) Vessel is thermally insulated thus no transfer of heat can take place

Heat transferred is:

Q=0

(b)Work done on system is:

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

(c) We know

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

From above parts 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

And

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Question 2

Figure (26.E1) shows a paddle wheel coupled to a mass of 12kg through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity 4200J/K kept in an adiabatic container. Consider a time interval in which the 12kg block falls slowly through 70cm. (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.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Solution 2

(a) Heat transfer does not take place but work is done which is then converted into heat.

Thus, heat given to liquid Q=0

(b) Work done on liquid,

W=P.E = mgh

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

= 84J

(c) We know

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Question 3

A 100kg block is started with a speed of 2.0m/s 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 from a frame of reference moving at 2.0m/s along the initial velocity of the block. As seen from its frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 20m/s .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.

Solution 3

We know

dQ=dU+dW

dU=-dW [As dQ=0]

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

dU=200J 

Question 4

Calculate the change in internal energy of a gas kept in a rigid container when 100J of heat is supplied to it.

Solution 4

We know,

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=100J

Question 5

The pressure of a gas changes linearly with volume from 10kPa, 200cc to 50kPa, 50cc. (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?

Solution 5

(a) Work done on gas is given as:

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

dW=-4.5J

(b) We know,

dQ=dU+dW

dU=-dW [As dQ=0]

=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 the gas?

Solution 6

According to question:

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Thus, change in volume is 0 i.e. H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

And work done by gas is given as:

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

W=0

Question 7

Figure (26-E2) shows three paths through which a gas can be taken from the state A to the B. Calculate the work done by the gas in each of the three paths.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Solution 7

(a) For AB path:

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=0.30J

b) For ACB path:

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=0.45J

c) For ADB path:

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

= 0.15J

Question 8

When a system is taken through the process abc shown in figure (26-E3), 80J of heat is absorbed by the system and 30J of work is done by it. If the system does 10J of work during the process adc, how much heat flows into it during the process?

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Solution 8

We know,

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Or H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=50J

And for ADC path

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=60J

Chapter 4 - Laws of Thermodynamics Exercise 63

Question 9

50cal of heat should be supplied to take a system from the state A to the state B through the path ACB as shown in figure (26-E4). Find the quantity of heat to be supplied to take it from A to B via ADB.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Solution 9

For ACB path

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

And H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Also We know,

dQ=dU+dW

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=210-10

dU=200J ____1

Now, for ADB path

dU = 200J [from 1]

andH-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=31J

Also we know,

dQ=dU+dW = 200+31

=231J or 55cal

Question 10

Calculate the heat absorbed by a system in going through the cyclic process shown in figure (26-E5).

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Solution 10

Heat absorbed is in form of work done which in this case is area covered in graph.

Heat = Area of circle

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=31.4J

Question 11

A gas is taken through a cyclic process ABCA as shown in figure, (26-E6). If 2.4 cl of heat is given in the process, what is the value of J?

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics

 

Solution 11

Figure shows cyclic process thus dU=0.

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

dW=10J

Also we know

dQ =dU+dW

2.4J=10

J=4.17J/cal

Question 12

A substance is taken through the process abc as shown in figure (26-E7). If the internal energy of the substance increases by 5000J and a heat of 2625cal is given to the system, calculate the value of J.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Solution 12

According to graph,

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=6000J

Also we know,

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

2625J=6000+5000

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Question 13

A gas is taken along the path AB as shown in figure (26-E8). If 70 cal of heat is extracted from the gas in the process, calculate the change in the internal energy of the system.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics

Solution 13

According to figure,

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=52.5J

And

dQ =-70cal

=-294J

Also we know

dQ = dU+dW

dU=dQ-dW

=-294-52.5

dU=-346.5J

Question 14

The internal energy of a gas is given by U=1.5pV. It expands from 100cm3 to 200cm3 against a constant pressure of 1.0×106 Pa Calculate the heat absorbed by the gas in the process.

Solution 14

We know

dU=UPdV

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

dU=15J

and dW =PdV = H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=10J

Also,

dQ=dU+dW

=15+10

dQ=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, 10J of heat is supplied and the piston is found to move out 10cm. Find the increase in the internal energy of the gas. The area of cross-section of the cylinder =4cm2 and the atmospheric pressure=100kPa.

Solution 15

We know

dW=PdV

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

dW=4J

AlsodQ = dV+dW

dU=dQ-dW

=10-4

=6J

Question 16

A gas is initially at a pressure of 100kPa and its volume is 2.0m3. Its pressure is kept constant and the volume is changed from 2.0m3 to 2.5m3. Its volume is now kept constant and the pressure is increased from 100kPa to 200kPa. The gas is brought back to 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?

Solution 16

(a) According to question

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

AndH-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

According to figure,

Area AC > Area AB

Heat is extracted from the system.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

 

b) Heat = Area ABC

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=25000J

Question 17

Consider the cyclic process ABCA, shown in figure (26-E9), performed on a sample of 2.0mole of an ideal gas. A total of 1200J of heat is withdrawn from the sample in the process. Find the work done by the gas during the part BC.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Solution 17

It is a cyclic process thus,

dU=0

and

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Also we know

dQ=dU+dW

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Question 18

Figure (26-E10) shows the variation in the internal energy U with the volume V of 2.0 mole of an ideal gas in a cyclic process abcda. The temperature of the gas at b and c are 500k and 300k respectively, Calculate the heat absorbed by the gas during the process.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Solution 18

It is a cyclic process H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Also H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Also dQ=dU+dW

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

dQ=2305.3J

Question 19

Find the change in the internal energy of 2kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J/kg-K and its densities at 0H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of ThermodynamicsC to 4°C are 999.9kg/m3 and 1000kg/m3 respectively. Atmospheric pressure=106Pa.

Solution 19

We know

dQ=dU+dW

dU=dQ-dW

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

dU=(33600-0.02)J

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 100kPa. The density of steam=0.6kg/m3. Specific heat capacity of water =4200J/kg-°C and the latent heat of vaporization of water=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Solution 20

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

dW=1699J

and H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

=29200

Also we know,

dQ=dU+dW

dU=dQ-dW

=29200-1699=27501

Chapter 4 - Laws of Thermodynamics Exercise 64

Question 21

Figure (26-E11) shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5nRT. 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?

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Solution 21

(a) The wall is fixed and cannot be moved. Thus,

dU=0 and dQ=0

and dW=dQ-dU=0

By ideal gas equation

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Also H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Now internal energy

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

And, U=1.5nRT

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Also

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

(b) Final pressure of left is H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics and right is H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

so,

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

(c) Temperature is given by equation 3

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

(d) Loss of heat is given as:

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

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 contains one mole of an ideal gas (U=1.5nRT) 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, (b) the heat given to the gas in the right part and (e) the increase in the internal energy of the gas is the left part.

Solution 22

(a) According to figure, wall is fixed

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

On left part

b) H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics (for left) and H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics (for right)

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

And H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

Where, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

(c) No. of mole H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

(d) For right part

dQ=dU+dW

dQ=dU  [As dW=0]

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics 

(e) dQ =-dU

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Laws Of Thermodynamics