Animated Solution for Physics - Thermodynamics: Comprehension Passage

The Hidden Identity of $X$

The passage presents us with a seemingly complex expression for a thermodynamic quantity $X$:

It also states that the infinitesimal heat absorbed is given by $T\Delta X$. Let's unravel this mystery. If we take the differential of $X$, we get:

Multiplying the entire equation by the temperature $T$, we obtain:

Recognizing that for an ideal monatomic gas, $dU=\frac{3}{2}RdT$ and from the ideal gas law $P=\frac{RT}{V}$, the equation beautifully transforms into:

This is the exact mathematical definition of entropy ($dS=\frac{dQ}{T}$). Thus, $X$ is simply the entropy of the gas! While this is a fascinating piece of physics trivia, the beauty of this problem is that we can solve it entirely using the First Law of Thermodynamics without ever needing to calculate $X$ directly. We are also given a crucial constraint: $P_0{V}_0=\frac{1}{3}R{T}_0$.

The PV Diagram Journey

Let's analyze the first cycle shown in the PV diagram.

Process $1\to 2$: This is a horizontal line, meaning it's an isobaric process at a constant pressure of $P_0$. The volume expands from $V_0$ to $2V_0$.

The work done is simply the area under the curve:

The change in internal energy for any ideal gas process can be calculated directly from the state coordinates using $\Delta U=\frac{f}{2}\Delta (PV)$. For a monatomic gas ($f=3$):

By the First Law, the heat absorbed is: This perfectly matches option (U) in List-II.

Process $2\to 3$: This is a vertical line, representing an isochoric process at a constant volume of $2V_0$. Since the volume doesn't change, the work done is strictly zero ($W_{2\to 3}=0$).

The change in internal energy is:

Total Process $1\to 2\to 3$: Summing it all up: - Total Work: $W=\frac{1}{3}R{T}_0+0=\frac{1}{3}R{T}_0$ (Matches Q) - Total $\Delta U=\frac{1}{2}R{T}_0+\frac{1}{2}R{T}_0=R{T}_0$ (Matches R) - Total Heat: $Q=W+\Delta U=\frac{4}{3}R{T}_0$ (Matches S)

Therefore, the correct mapping is I $\to$ Q, II $\to$ R, III $\to$ S, IV $\to$ U, which corresponds to option (C).

The TV Diagram Expedition

Now, let's shift our focus to the TV diagram.

Process $1\to 2$: This is a horizontal line on a TV diagram, which means it's an isothermal process. But what is the temperature? Using the ideal gas law at state 1 ($P_0,V_0$), we know $T_1=\frac{P_0{V}_0}{R}$. Since $P_0{V}_0=\frac{1}{3}R{T}_0$, the temperature is $T_1=\frac{T_0}{3}$.

The work done in an isothermal expansion is:

Because the temperature is constant, the change in internal energy is zero ($\Delta U_{1\to 2}=0$). Consequently, all the heat absorbed goes into doing work: This matches option (P).

Process $2\to 3$: This is a vertical line, meaning the volume is constant at $2V_0$ (isochoric). Again, the work done is zero ($W_{2\to 3}=0$).

The temperature increases from $\frac{T_0}{3}$ to $T_0$. The change in internal energy is:

Total Process $1\to 2\to 3$: Summing it all up: - Total Work: $W=\frac{1}{3}R{T}_0\ln 2+0=\frac{1}{3}R{T}_0\ln 2$ (Matches P) - Total $\Delta U=0+R{T}_0=R{T}_0$ (Matches R) - Total Heat: $Q=W+\Delta U=\frac{1}{3}R{T}_0\ln 2+R{T}_0=\frac{1}{3}R{T}_0(3+\ln 2)$ (Matches T)

Therefore, the correct mapping is I $\to$ P, II $\to$ R, III $\to$ T, IV $\to$ P, which corresponds to option (D).

This problem is a masterclass in reading thermodynamic graphs. Whether it's a PV diagram or a TV diagram, the core principles remain the same. Identify the process, apply the specific formulas for Work and Internal Energy, and let the First Law of Thermodynamics guide you to the Heat absorbed. Don't let intimidating formulas like the one for entropy ($X$) distract you from the fundamental physics at play!