Q19. Second Law of Thermodynamics

1. An ideal gas, consisting of n moles, undergoes an

irreversible process in which the temperature has the same

value at the beginning and end. If the volume changes

from Vi to Vf , the change in entropy is given by:

1. n R ( Vf – Vi )

2. n R ln( Vf – Vi )

3. n R ln( Vi / Vf )

4. n R ln( Vf / Vi ) 5. none of the above (entropy can't be

calculated for an irreversible process)

Temperature has the same value at the beginning and end

Initial & final states can be related by an isothermal reversible process.

Ideal gas 2

fU n R T

lnf fV V

f

iVi Vi

VdQ d W P nRS d V d V nR

T T T V V

0dU dQ dW Isothermal

P V n R T

2. The temperature of n moles of a gas is increased from Ti to Tf at

constant pressure. If the molar specific heat at constant

pressure is Cp and is independent of temperature, then change in

the entropy of the gas is:

1. n Cp ln( Tf / Ti )

2. n Cp ln( Ti / Tf )

3. n Cp ln( Tf – Ti )

4. n Cp ln(1 – Ti / Tf )

5. n Cp ( Tf – Ti )

dQS

T

f

i

T

P

T

n C dT

T ln f

Pi

Tn C

T

3. Consider the following processes: The temperature of two

identical gases are increased from the same initial temperature

to the same final temperature. Reversible processes are

used. For gas A the process is carried out at constant volume

while for gas B it is carried out at constant pressure. The

change in entropy:

1. is the same for A and B

2. is greater for A

3. is greater for B

4. is greater for A only if the initial temperature is low

5. is greater for A only if the initial temperature is high

dQS

T

f

i

T

T

C dT

T

P VC C P VS S

4. A Carnot heat engine runs between a cold reservoir at

temperature TC and a hot reservoir at temperature TH.

You want to increase its efficiency. Of the following, which

change results in the greatest increase in efficiency?

The value of T is the same for all changes.

1. Raise the temperature of the hot reservoir by T

2. Raise the temperature of the cold reservoir by T

3. Lower the temperature of the hot reservoir by T

4. Lower the temperature of the cold reservoir by T

5. Raise the temperature of the hot reservoir by

(1/2)T and lower the temperature of the cold reservoir by

(1/2)T

1 C

H

Te

T

1 2

1C

H H

Te T T

T T

2

1CH C

H H

Te T T

T T

2

1

H

e TT

3 2C

H

Te T

T

5 2

1 1

2C

H H

Te T

T T

4

1

H

e TT

2

1

2H C

H

T TT

T

1

H

TT

5. A perfectly reversible heat pump with a coefficient of

performance of 14 supplies energy to a building as heat to

maintain its temperature at 27°C. If the pump motor does

work at the rate of 1 kW, at what rate does the pump supply

energy to the building?

1. 15 kW

2. 14 kW

3. 1 kW

4. 1.07 kW

5. 1.02 kW

CQCOP

W

14 1 14CdQkW kW

dt

Energy pump rate into building :

14 1 1 15H CdQ dQW kW kW kW

dt d t

6. A Carnot heat engine and an irreversible heat engine both

operate between the same high temperature and low

temperature reservoirs. They absorb the same heat from the

high temperature reservoir as heat. The irreversible engine:

1. does more work

2. rejects more energy to the low temperature reservoir as

heat

3. has the greater efficiency

4. rejects less energy to the low temperature reservoir as

heat 5. cannot absorb the same energy from the high

temperature reservoir as heat without violating the second law of thermodynamics

H HQ Q

2nd law has nothing to say about the magnitude of QH

H CCarnot irrev

H H H

W T T We e

Q T Q

W W

C H

C H

Q Q W

Q Q W

C CQ Q

5 is false

2 is true4 is false

3 is false

1 is false

7. Twenty-five identical molecules are in a box. Microstates

are designated by identifying the molecules in the left and

right halves of the box. The Boltzmann constant is 1.38

10–23 J / K. The entropy associated with the configuration

for which 22 molecules are in the left half and 3 molecules are

in the right half is on the order of:

1. 10–23 J/K

2. 10–20 J/K

3. 103 J/K

4. 1022 J/K

5. 10–10 J/K

25! 25 24 232300

22! 3! 3 2

–23 231.38 10 / ln 230ln 0 10JS k K

8. Let k be the Boltzmann constant. If the thermodynamic

state of gas at temperature T changes isothermally and

reversibly to a state with three times the number of

microstates as initially, the energy input to gas as heat is :

1. Q = 0

2. Q = 3 k T

3. Q = –3k T

4. k T ln 3

5. – k T ln 3

ln ln ln ln 3ff i

i

Q T S T k T k k T

isothermal