Animated Solution for Physics - Thermodynamics: One mole of a monatomic ideal gas undergoes the cyclic process J $\to$ K $\to$ L $\to$ M $\to$ J, as shown in the P-T diagram. Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [R is the gas constant.]

Thermodynamics is often perceived as a labyrinth of equations, but at its core, it is a beautiful story of energy transformation. In this problem, we are presented with a classic scenario: one mole of a monatomic ideal gas undergoing a cyclic process. The journey is mapped out on a Pressure-Temperature (P-T) diagram, forming a neat rectangle through states $J\to K\to L\to M\to J$. Our mission is to dissect this cycle, analyze each leg, and match the thermodynamic quantities to their correct values.

Before diving into calculations, let's orient ourselves. A P-T diagram is incredibly revealing. A horizontal line indicates a process where pressure is constant—an isobaric process. A vertical line indicates a process where temperature is constant—an isothermal process.

Our cycle consists of two isobaric processes ($J\to K$ and $L\to M$) and two isothermal processes ($K\to L$ and $M\to J$). Since the gas is monatomic, we know its molar heat capacity at constant volume is $C_V=\frac{3}{2}R$.

Isobaric Expansion

Let's begin with the first leg, from state $J$ to state $K$. The graph shows a horizontal line at a constant pressure $P_0$. The temperature, however, increases from $T_0$ to $3T_0$.

We are asked to find the change in internal energy ($\Delta U_{JK}$). For any ideal gas, the change in internal energy depends solely on the change in temperature, governed by the equation:

Substituting our known values ($n=1$, $C_V=\frac{3}{2}R$, and $\Delta T=3T_0-T_0=2T_0$), we get:

This perfectly matches option (3) in List-II. So, (Q) maps to (3).

Isothermal Compression

Next, the gas moves from $K$ to $L$. Here, the temperature is locked at $3T_0$, making it an isothermal process. The pressure increases from $P_0$ to $2P_0$.

We need to find the heat given to the gas ($Q_{KL}$). The First Law of Thermodynamics states that $Q=\Delta U+W$. Since the process is isothermal, the temperature doesn't change, meaning $\Delta U_{KL}=0$. Therefore, the heat exchanged is exactly equal to the work done: $Q_{KL}=W_{KL}$.

For an isothermal process, the work done can be expressed in terms of pressure as:

Plugging in the values for this specific leg ($T=3T_0$, $P_i=P_0$, $P_f=2P_0$):

This matches option (5). Thus, (R) maps to (5).

Isothermal Expansion

Let's skip ahead to the final leg, $M\to J$, to find its change in internal energy. Just like $K\to L$, this is a vertical line on the P-T diagram, meaning the temperature is constant at $T_0$.

As we established earlier, if there is no change in temperature ($\Delta T=0$), there is absolutely no change in the internal energy of an ideal gas.

This straightforward deduction matches option (2). Therefore, (S) maps to (2).

Total Work Done

Finally, we must calculate the total work done in the complete cyclic process. This is the algebraic sum of the work done in all four individual legs:

We already know $W_{KL}=-3R{T}_0\ln 2$. Let's find the work for the remaining three processes.

For the isobaric processes ($J\to K$ and $L\to M$), the work done is simply $W=nR\Delta T$:

Notice the beautiful symmetry! The work done during the isobaric expansion is perfectly canceled out by the work done during the isobaric compression. $W_{JK}+W_{LM}=0$.

Now, let's calculate the work for the final isothermal process, $M\to J$:

Adding the non-zero components together yields the total work:

This matches option (4). So, (P) maps to (4).

By systematically breaking down the cycle and applying fundamental thermodynamic principles, we have successfully decoded the entire problem. The final mapping is P $\to$ 4, Q $\to$ 3, R $\to$ 5, S $\to$ 2.