q18. first law of thermodynamics

Q18. First Law of Thermodynamics

1. A quantity of an ideal gas is compressed to half its initial

volume. The process may be adiabatic, isothermal or

isobaric. Rank those three processes in order of the work

required of an external agent, least to greatest.

1. adiabatic, isothermal,

isobaric

2. adiabatic, isobaric,

isothermal

3. isothermal, adiabatic,

isobaric

4. isobaric, adiabatic,

isothermal

5. isobaric, isothermal,

adiabatic

AdiabaticIsothermalIsobaric

P

2P

2 P

2. When an ideal gas undergoes a slow isothermal expansion:

1. (work done by the gas) = (energy absorbed as heat )

2. (work done by the environment ) = (energy absorbed as heat )

3. (increase in internal energy ) = ( heat absorbed )

4. (increase in internal energy ) = ( work done by the gas )

5. ( increase in internal energy ) = ( work done by the

environment )

Isothermal dU = 0 d Q = d W

Expansion d W = P d V > 0 (Work done by gas)

d Q > 0 Heat absorbed

dU dQ dW 3

2U n R T

1. (work done by the gas) = (energy absorbed as heat ) d W =

d Q

2. (work done by the environment ) = (energy absorbed as heat ) d W =

d Q

3. (increase in internal energy ) = ( heat absorbed ) d U = d Q

4. (increase in internal energy ) = ( work done by the gas ) d U = d W

5. ( increase in internal energy ) = ( work done by the environment ) d U = d W

3. An ideal gas of N monatomic molecules is in thermal

equilibrium with an ideal gas of the same number of diatomic

molecules and equilibrium is maintained as temperature is

increased. The ratio of the changes in the internal energies

ΔEdia / Emon is :

1. 1/2

2. 3/5

3. 1

4. 5/3

5. 2

2

fU N k T

2

fU N k T

5

3dia dia

mon mon

U f

U f

4. The pressure of an ideal gas of diatomic molecules is doubled

by halving the volume. The ratio of the new internal energy

to the old, both measured relative to the internal energy at 0

K, is:

1. 1/4

2. 1/2

3. 1

4. 2

5. 4

P V n R T2

fU N k T

22

VP P V

T remains the same.

So does U.

5. When work W is done on an ideal gas of N diatomic

molecules in thermal isolation the temperature increases by

1. W / 2 N k

2. W / 3 N k

3. 2 W / 3 N k

4. 2 W / 5 N k

5. W / N k

dU dQ dW

5

2U N k T

2

5T W

N k

Diatomic molecules :

Thermal isolation : dQ = 0 U W Work W done on gas:

6. When work W is done on an ideal gas of diatomic molecules

in thermal isolation the increase in the total rotational energy

of the molecules is:

1. 0

2. W / 3

3. 2 W / 3

4. 2 W / 5

5. W

dU dQ dW

2

5rotU U

2

5rotU W

Diatomic molecules :

Thermal isolation : dQ = 0 U W

3

5tranU U

Work W done on gas:

7. The pressure of an ideal gas is doubled during a process in

which the energy given up as heat by the gas equals the work

done on the gas. As a result, the volume is:

1. doubled

2. halved

3. unchanged

4. need more information to

answer

5. nonsense, the process is

impossible

dU dQ dW

T const P V n R T const

0U Q W 2

fU n R T

2P P 2

VV

Heat Q given up by the gas equals work W done on the gas

8. The temperature of n moles of an ideal monatomic gas is

increased by T at constant pressure. The energy Q

absorbed as heat, change Eint in internal energy, and work W

done by the environment are given by:

1. Q = (5/2)nRT, Eint = 0, W = –nRT

2. Q = (3/2)nRT, Eint = (5/2)nRT, W = –

(3/2)nRT

3. Q = (5/2)nRT, Eint = (3/2)nRT, W = nRT

4. Q = (3/2)nRT, Eint = 0, W = nRT

5. Q = (5/2)nRT, Eint = (3/2)nRT, W = –nRT

51

2 2P

fC nR nR

3

2U n R T

n RW P V P T n R T

P

5

2PQ C T n R T