“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 ΔT º C. If the
specific heats of the gas at constant volume and constant pressure are Cv and Cp respectively, the increase in internal energy of the
gas is
(a) m Cp
ΔT
(b) m (Cp – Cv)
ΔT
(c) m Cv
ΔT
(d) m (Cp + Cv)
ΔT
(e) m (Cp + Cv)
ΔT/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 (Cv). 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 Cv.
Therefore, the
increase in internal energy (ΔU)
of m kg of the gas
when its temperature is raised through ΔT
º C is given by
ΔU = m Cv ΔT
(2) In the above question the work done by the gas is
(a) mCpΔT
(b) m(Cp – Cv)ΔT
(c) mCvΔT
(d) m(Cp + Cv)ΔT
(e) Zero
Keeping the pressure constant, when the temperature of m kg of the gas is raised through ΔT
º C, the total energy required is mCpΔT. Out of this the energy used in increasing
the internal energy of the gas is mCvΔT. The energy spent for doing work is
therefore given by the difference between the two values and is equal to m(Cp – Cv)ΔT.
(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 = (Q1
– Q2)/Q1 = (T1 –
T2)/T1
where Q1 is the quantity
of heat absorbed from the source at the source temperature T1 and Q2
is the quantity of heat liberated to the sink at the sink temperature T2.
The efficiency η1 when the source and sink are at 400 K and 300 K
respectively is given by
η1 = (400 – 300)/400 = ¼
The efficiency η2 when the source and sink are at 400 K and 200 K
respectively is given by
η2 = (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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