Thermodynamics – Two Multiple Choice Questions

Often simple questions on thermodynamics may cause unexpected confusion. The reason for the confusion is usually the lack of your understanding of the fundamentals. If you don’t have any confusion in respect of the following questions, well and good!

(1) A sample of an ideal gas initially having internal energy U₁ and pressure P₁ expands adiabatically and performs work W. Heat energy Q is then added to the gas at constant volume so that its pressure is increased to the initial value P₁. As a result of the above processes, the internal energy of the gas

(a) decreases by Q – W

(b) increases by Q – W

(c) decreases by Q

(d) increases by Q

(e) remains unchanged

Since the gas expands adiabatically and outputs mechanical energy, the internal energy of the gas is decreased. [Remember that during adiabatic process there is no heat transfer between the gas and the surroundings]. On adding heat energy to the gas at constant volume the internal energy of the gas is increased. There is no work involved since the volume is constant (isochoric process).

Evidently the internal energy of the gas increases by Q – W.

[You will obtain the answer from the mathematical statement of the 1^st law of thermodynamics: ∆Q = ∆U + ∆W where ∆Q is the heat energy supplied to the system by the surroundings, ∆W is the work done by the system on the surroundings and ∆U is the increase in the internal energy of the system].

(2) In the given PV diagram, I is the initial state and F is the final state. The gas goes from I to F by (i) IAF (ii) IBF (iii) ICF. The heat absorbed by the gas is

(a) the same in all three processes

(b) the same in (i) and (ii)

(c) greater in (i) than in (ii)

(d) the same in (i) and (iii)

(e) greater in (iii) than in (ii)

This question appeared in Kerala engineering entrance 2009 question paper.

We have ∆Q = ∆U + ∆W

Since the same initial point (I) and the same final point (F) are given for all the three paths, the change in the internal energy (∆U) is the same for all the three processes. During the path IAF the net work done by the gas (∆W) is positive since the work done on the gas during the compression IA (given by the area under the line IA) is less than the work done by the gas during its expansion AF. No work is involved during the process IBF since the volume is constant (isochoric process). During the path ICF the net work done (∆W) by the gas is negative since the work done by the gas during its expansion IC is less than the work done on the gas during its compression CF.

Therefore, the heat absorbed by the gas (∆Q) is greater in (i) than in (ii).

You will find useful multiple choice questions with solution at apphysicsresources as well as at physicsplus.