“This time, like
all times, is a very good one if we but know what to do with it”

– Ralph Waldo
Emerson

You will find many practice questions (with solution) on thermodynamics
on this site. You can access them by clicking on the label ‘thermodynamics’
below this post. Today we shall discuss a few more multiple choice practice
questions in this section for the benefit of AP Physics B aspirants.

(1) Keeping the pressure constant, the temperature of

*m*kg of a gas is raised through Δ*T*º C. If the specific heats of the gas at constant volume and constant pressure are*C*_{v}and*C*_{p}respectively, the increase in internal energy of the gas is
(a)

*m**C*_{p}Δ*T*
(b)

*m*(*C*_{p}*–**C*_{v}) Δ*T*
(c)

*m**C*_{v}Δ*T*
(d)

*m*(*C*_{p}*+**C*_{v}) Δ*T*
(e)

*m*(*C*_{p}*+**C*_{v}) Δ*T*/2
If the temperature
increase occurs at constant volume, the entire energy supplied to the gas is
used up in increasing the internal energy of the gas. The energy required for
increasing the temperature of one kilogram of the gas through 1º C at constant
volume is the

*specific heat*of the gas*at constant volume*(*C*_{v}). When the temperature increase occurs at constant pressure, the gas has to expand and therefore has to do external work. The specific heat of the gas at constant pressure is equal to the energy supplied for increasing the temperature of one kilogram of the gas through 1º C at constant pressure. But the increase in the internal energy of 1 kg of the gas on heating through 1º C in this case also is*C*_{v}.
Therefore, the
increase in internal energy (Δ

*U*) of*m*kg of the gas when its temperature is raised through Δ*T*º C is given by
Δ

*U = m**C*_{v}Δ*T*
(2) In the above question the work done by the gas is

(a)

*m**C*_{p}Δ*T*
(b)

*m*(*C*_{p}*–**C*_{v})Δ*T*
(c)

*m**C*_{v}Δ*T*
(d)

*m*(*C*_{p}*+**C*_{v})Δ*T*
(e)

*Zero*
Keeping the pressure constant, when the temperature of

*m*kg of the gas is raised through Δ*T*º C, the total energy required is*m**C*_{p}Δ*T*. Out of this the energy used in increasing the internal energy of the gas is*m**C*_{v}Δ*T*. The energy spent for doing work is therefore given by the difference between the two values and is equal to*m*(*C*_{p}*–**C*_{v})Δ*T*.
(3) A sample of gas contained in a cylinder undergoes a cyclic process
shown by the adjoining PV diagram. Among the following statements which one is
correct?

(a) Work is done on the gas during the process
shown by AB

(b) Work done by the gas during the cycle ABCA is
negative

(c) BC represents an isochoric process

(d) CA represents an isobaric process

(e) Work done on the gas during the process CA is
zero

During the process shown by AB the gas

*expands*. Therefore work is done*by*the gas (and not*on*the gas). Statement (a) is therefore incorrect.
Since the cycle ABCA is clockwise, work done by
the gas is positive. Statement (b) is therefore incorrect.

[Work done

*by the gas*during its expansion represented by the area under the curve AB is greater than the work done*on the gas*during its contraction represented by the area under the curve BC]
During the process BC the

*volume of the gas changes*and hence the process is not isochoric.
[Isochoric process is also known as

*isovolumetric*process]
During the process CA the

*pressure of the gas changes*and hence the process is not isobaric.
Work done on the gas uring the process CA is zero
since CA represents an

*isochoric*process. Therefore option (e) is correct.
[Note that work is done only if volume changes].

(4) A Carnot engine
operates using a high temperature source at 400 K and a low temperature sink at
300 K. How much more efficient will this engine be if the temperature of the
sink is reduced to 200 K?

(a) Twice as efficient

(b) Three times as efficient

(c) Four times as efficient

(d) Five times as efficient

(e) Six times as efficient

Efficiency (η) of Carnot engine is given by

*η*= (

*Q*

_{1 }–

*Q*

_{2})/

*Q*

_{1}= (

*T*

_{1 }–

*T*

_{2})/

*T*

_{1}where

*Q*

_{1 }is the quantity of heat absorbed from the source at the source temperature

*T*

_{1}and

*Q*

_{2 }is the quantity of heat liberated to the sink at the sink temperature

*T*

_{2}.

The efficiency

*η*_{1}when the source and sink are at 400 K and 300 K respectively is given by*η*

_{1}= (400 – 300)/400 = ¼

The efficiency

*η*_{2}when the source and sink are at 400 K and 200 K respectively is given by*η*

_{2}= (400 – 200)/400 = ½

The engine is therefore

*twice*as efficient [Option (a)].
(5) A fixed mass of gas does 25 J of work on its surroundings and transfers
15 J of heat to the surroundings. The internal energy of the gas

(a) decreases by 15 J

(b) increases by 15 J

(c) remains unchanged

(d) decreases by 40 J

(e) increases by 40 J

The internal energy of the gas decreases when it
does work on the surroundings and also when it transfers heat to the
surroundings. The total decrease in the internal energy of the gas is therefore
equal to 25 J + 15 J = 40 J [Option (d)].

You will find similar questions with solution here.

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