The problem that arose my confusion:

One mole of a monatomic ideal gas undergoes a cyclic process as shown in the figure (where $V$ is the volume and $T$ is the temperature). Which of the statements below is (are) true?

enter image description here

(A) Process I is an isochoric process

(B) In process II, gas absorbs heat

(C) In process IV, gas releases heat

(D) Processes I and III are not isobaric

The correct options are

(B) (C) (D),

and I understand

how (A) is false.

My confusion: I was having issues when dealing with the thermodynamical process that the gas undergoes during process II. It seems that statement (B) need not necessarily be true since II can be a free expansion process. If II allows for heat to flow in and out of the system (so now it is an Isothermal expansion), then (B) is a true statement. How does one resolve such a conflict?

  • 1
    $\begingroup$ Since the temperature is shown varying with volume at every point through the process, the process must be reversible. Otherwise, the temperature would not be well-defined through the process, since the system would not be close to equilibrium. Free expansion is an irreversible process, so this diagram would not correspond to that. $\endgroup$ – Chet Miller Mar 7 '19 at 12:34
  • $\begingroup$ Thanks, this is what was confusing me. Before, I wasn't able to to convince myself that this is a reversible process. $\endgroup$ – Apekshik Panigrahi Mar 11 '19 at 13:22

Well, since it is an ideal gas, the internal energy is a function only of the temperature T. Since process $II$ is isothermal, there is no net change in internal energy and thus:

$$\Delta U = \Delta W + \Delta Q = 0$$

With this sign convention, $Q$ is the heat supplied to the system and the work done by the system to the surroundings is $dW = -pdV$.

Implying $\Delta W = - \Delta Q$. Now, use the ideal gas law $PV = nRT$ to compute work:

$$\Delta W = \int_{II} dW = \int_{II} -pdV = \int_{V_{II}}^{V_{III}} -\frac{nRT}{V}dV = -nRT \ln(\frac{V_{III}}{V_{II}}) $$

Now, since $$\Delta Q = -\Delta W = nRT\ln(\frac{V_{III}}{V_{II}})$$

and $V_{III}$ is greater than $V_{II}$, $\Delta Q > 0$, meaning that the system absorbs heat, as stated in option (B)

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  • $\begingroup$ Well, can't the process be adiabatic and isothermal as well (since it hasn't been mentioned in the problem)? So then the process II becomes a free expansion doesn't it? $\endgroup$ – Apekshik Panigrahi Mar 7 '19 at 8:31

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