Isothermal process of free expansion?

Question

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?

(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

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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?

Answer 1

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)