# Isothermal process of free expansion?

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

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?

• 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. Mar 7, 2019 at 12:34
• Thanks, this is what was confusing me. Before, I wasn't able to to convince myself that this is a reversible process. Mar 11, 2019 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)

• 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? Mar 7, 2019 at 8:31