AP Physics B – Multiple Choice Questions on Thermodynamics

The essential things to be remembered in thermodynamics were given in the post dated 21^st May 2008.

We will now discus some typical multiple choice practice questions related to thermodynamics. Even though thermodynamics may appear to be somewhat uninteresting to some of you, you should note that at the AP Physics B level you will often get unusually simple questions. So, cheer up and go through these questions and after your own trial, see their solution:

(1) In the PV diagram shown, points A and B represent two states of a given mass of an ideal gas. This gas is taken from state A to state B in two different ways:

(i) An isobaric process first and an isochoric (constant volume) process next.

(ii) An isochoric process first and an isobaric process next.

If the work done by the gas in cases (i) and (ii) are respectively W₁ and W₂, pick out the most suitable statement:

(a) W₁> W₂

(b) W₂> W₁

(c) W₁= W₂

(d) W₁= W₂/2

(e) W₁= W₂/3

In both cases the gas expands and therefore work is done by the gas. You should remember that no work is done in the case of an isochoric process. So, no work is done in the processes indicated by PB and AQ (Fig.).

Work is done by the gas during the processes indicated by AP and QB. Since the work done is equal to the area under the PV curve, more work is done during the process AP. Therefore, W₁> W₂.

(2) In the cyclic process on an ideal gas shown in the adjoining PV diagram, what is the net work done on the gas during the cycle?

(a) 2PV

(b) 4PV

(c) – 2PV

(d) – 4PV

(e) zero

This is a simple question of the type often asked. You should note that in a cyclic process the gas is taken through a cycle of operations and brought back to the original state. If the cycle is clockwise as in the present case, work is done by the gas. So the work done on the gas is negative. Further, the area enclosed by the closed curve gives the work done during one cycle.

Therefore, work done on the gas,

W = – area of the rectangle APBQA

= – 2P×2V = – 4PV.

(3) A sample of gas (assumed to be ideal) is adiabatically compressed to have its volume reduced to 20% of its initial volume. If the internal energy of the gas is increased by 200 J, the work done on the gas must be

(a) 40 J

(b) 100 J

(c) 200 J

(d) 400 J

(e) zero

Since the process is adiabatic, no heat energy flows from the surroundings to the gas or from the gas to the surroundings. So, the entire work done in compressing the gas is utilised in increasing the internal energy of the gas. The work done on the gas is therefore equal in value to the increase in internal energy, which is 200 J.

(4) The figure shows the variation of the pressure of a gas with temperature (and not volume) of a given mass of an ideal gas subjected to various processes starting from the state represented by point A. Which graph represents an isochoric process (a process in which volume is constant)?

(a) AB

(b) AC

(c) AD

(d) AE

(e) AF

Some of you may be baffled on seing this simple question since most of you are accustomed to PV diagrams only. The required graph is AB [Option (a)] since the pressure of a given mass of gas is directly proportional to its absolute temperature when its volume is constant (in accordance with Charles law). This type of pressure-volume relationship exists only in the case of the process shown by graph AB.

(5) The volume of a gaseous sample is reduced from V₁ to V₂ in three different ways:

(i) isobaric process (ii) isothermal process and (iii) adiabatic process. Then the work done by the gas is

(a) minimum in the isothermal process

(b) maximum in the isothermal process

(c) minimum in the isobaric process

(d) maximum in the isobaric process

(e) maximum in the adiabatic process

You can easily obtain the answer by referring to the PV diagram showing the three processes (Fig.). Since the area under the PV curve gives the work done, the maximum work is for the isobaric (constant pressure) process indicated by the cuve AB. So the correct option is (d).

The isothermal curve is AC and the adiabatic curve is AD. (Note that the adiabatic curve is steeper than the isothermal curve. The slope of the adiabatic curve is γ times the slope of the isothermal curve where γ is the ratio of specific heats of the gas)

We will discuss more questions in this section in due course