Animated Solution for Physics - Thermodynamics: One mole of a monoatomic 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 ?

Thermodynamic graphs are like maps; they tell us exactly what a gas is experiencing at any given moment. In this problem, we are given a cyclic process plotted on a $T-V$ (Temperature-Volume) graph. To determine which statements are true, we need to translate the geometry of the lines into physical thermodynamic processes.

Processes II and IV

Let's start with the easiest features to identify: the horizontal lines.

In Process II, the graph moves horizontally to the right. A horizontal line on a $T-V$ graph means the temperature $T$ is constant. This defines an isothermal process. Because the graph moves to the right, the volume $V$ is increasing, making it an isothermal expansion.

For an ideal gas, the internal energy $U$ depends only on temperature. Since $\Delta T=0$, the change in internal energy $\Delta U=0$. The gas is expanding, so it does positive work ($W>0$). According to the First Law of Thermodynamics:

Since $\Delta U=0$ and $W>0$, it follows that $\Delta Q>0$. The positive sign indicates that the gas absorbs heat. Therefore, statement (b) is correct.

Similarly, Process IV is a horizontal line moving to the left. This is an isothermal compression ($T=\text{constant}$, $V$ decreases). Here, $\Delta U=0$ and $W<0$. Consequently, $\Delta Q<0$, meaning the gas releases heat. Thus, statement (c) is also correct.

Are they Isobaric?

Now, let's evaluate the slanted lines, Process I and Process III. Statement (d) claims they are not isobaric. To verify this, we must recall the condition for an isobaric (constant pressure) process on a $T-V$ graph.

From the ideal gas law, $PV=nRT$, we can rearrange it to solve for temperature:

If the pressure $P$ is constant, the term $\left(\frac{P}{nR}\right)$ is a constant slope. This equation takes the form $y=mx$, which represents a straight line that must pass exactly through the origin $(0,0)$.

If we look closely at the given graph and extend the lines for Process I and Process III backwards, they intersect the volume axis at a positive value, not the origin. Because they do not pass through the origin, the pressure is not constant. Therefore, they are not isobaric processes, making statement (d) correct.

Finally, let's quickly check statement (a), which suggests Process I is isochoric. An isochoric process means the volume is constant, which would appear as a perfectly vertical line on a $T-V$ graph. Since Process I is slanted and the volume is clearly changing, statement (a) is incorrect.

By systematically analyzing the geometry of the graph and applying the First Law of Thermodynamics, we confidently conclude that the correct statements are (b), (c), and (d).