In the reported figure, there is a cyclic process ABCDA on a sample of 1 mol of a diatomic gas. The

Q: Why is the work done in an adiabatic process dependent only on temperature?

In an adiabatic process, there is no heat exchange ($Q = 0$). According to the First Law of Thermodynamics, $W = -\Delta U$. Since the internal energy of an ideal gas depends only on its temperature ($\Delta U = nC_V\Delta T$), the work done is strictly determined by the temperature difference.

Q: Why are we calculating $W_{AD}$ instead of $W_{DA}$?

The options specifically ask us to evaluate the mathematical expression $W_{AD}$, which represents the work done if the system transitioned from state A to state D. Even though the actual cycle goes $D \rightarrow A$, we must evaluate the exact term presented in the options.

Q: Are $W_{AB}$ and $W_{CD}$ equal?

No. $W_{AB}$ and $W_{CD}$ represent work done during isothermal processes. Isothermal work is given by $W = nRT \ln(V_f/V_i)$. Since the temperatures $T_1$ and $T_2$ are different, and the volume ratios are generally different, the work done in these two processes will not be equal.

Q: What does the area enclosed by the cycle ABCDA represent?

The area enclosed by the closed loop on a $p-V$ diagram represents the net work done by the gas during one complete cycle. It is the algebraic sum of the work done in all four individual processes.

Q: Why is the parameter $\gamma$ important here?

The parameter $\gamma$ is the ratio of specific heats ($C_p/C_V$). It dictates the steepness of the adiabatic curve on the $p-V$ diagram and directly appears in the denominator of the adiabatic work formula, determining how much mechanical work is extracted for a given drop in temperature.

Question

In the reported figure, there is a cyclic process ABCDA on a sample of $1	ext{ mol}$ of a diatomic gas. The temperature of the gas during the process $A ightarrow B$ and $C ightarrow D$ are $T_1$ and $T_2$ ($T_1 > T_2$), respectively.

Choose the correct option out of the following for work done, if processes BC and DA are adiabatic.

Options:

(A) $W_{AB} = W_{DC}$

(B) $W_{AD} = W_{BC}$

(C) $W_{BC} + W_{DA} > 0$

(D) $W_{AB} < W_{CD}$

Sigma Percentile · Accepted Answer

### The Elegance of Adiabatic Work in a Cyclic Process

Imagine you are observing a gas trapped inside a cylinder, undergoing a perfectly orchestrated sequence of expansions and compressions. This is exactly what the cyclic process $ABCDA$ represents on our $p-V$ diagram. To solve this problem, we need to dissect the cycle into its fundamental thermodynamic components and analyze the work done during specific transitions.

### Analyzing the Setup

The problem states that the temperature of the gas during the process $A \rightarrow B$ is a constant $T_1$, and during $C \rightarrow D$ it is a constant $T_2$. This immediately tells us that these two paths are **isothermal processes**.

On the other hand, the processes connecting these isotherms, namely $B \rightarrow C$ and $D \rightarrow A$, are explicitly given as **adiabatic processes**. In an adiabatic process, the system is perfectly insulated; no heat enters or leaves the gas ($Q = 0$).

### The Master Equation for Adiabatic Work

When a gas expands or compresses adiabatically, the work it does comes entirely at the expense of its internal energy. According to the First Law of Thermodynamics ($Q = \Delta U + W$), since $Q = 0$, we have $W = -\Delta U$.

For an ideal gas, the change in internal energy is strictly a function of temperature: $\Delta U = nC_V\Delta T$. Using the relation $C_V = \frac{R}{\gamma - 1}$, we can write the master equation for adiabatic work as:

$$W = \frac{nR(T_i - T_f)}{\gamma - 1} = \frac{nR(T_f - T_i)}{1 - \gamma}$$

Notice something beautiful here: the work done in an adiabatic process depends *only* on the initial and final temperatures, regardless of the specific pressures or volumes involved!

### Evaluating the Specific Paths

The question asks us to evaluate the relationship between the work done in various paths. Let's look at option (b), which compares $W_{AD}$ and $W_{BC}$.

**1. Process $A \rightarrow D$:**
If the gas were to move from state $A$ to state $D$, it starts on the upper isotherm at temperature $T_1$ and ends on the lower isotherm at temperature $T_2$. Plugging these into our master equation:

$$W_{AD} = \frac{nR(T_D - T_A)}{1 - \gamma} = \frac{nR(T_2 - T_1)}{1 - \gamma}$$

**2. Process $B \rightarrow C$:**
Now, let's look at the actual forward path from $B$ to $C$. The gas starts at state $B$, which also lies on the upper isotherm $T_1$. It expands adiabatically to state $C$, which lies on the lower isotherm $T_2$. Using the exact same logic:

$$W_{BC} = \frac{nR(T_C - T_B)}{1 - \gamma} = \frac{nR(T_2 - T_1)}{1 - \gamma}$$

### The Final Conclusion

By simply comparing the two mathematical expressions we just derived, the conclusion is inescapable:

$$W_{AD} = W_{BC}$$

Because both adiabatic paths bridge the exact same two temperature reservoirs ($T_1$ and $T_2$), the change in internal energy is identical, and thus the work done is identical. This elegant symmetry confirms that option (b) is the correct answer.

Select Answer:

Visualized Solution

Visualizing the Cyclic Process

Work Done in an Adiabatic Process

Evaluating $W_{A D}$

Evaluating $W_{B C}$

Comparing the Work Done

The Way Forward: Isothermal Work

The Sigma Insight: Thermodynamic Processes

The Elegance of Adiabatic Work in a Cyclic Process

Analyzing the Setup

The Master Equation for Adiabatic Work

Evaluating the Specific Paths

The Final Conclusion

Similar Questions

The $p - V$ diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process $C D$ is (use, $γ = 1.4$ )

One mole of an ideal gas in initial state $A$ undergoes a cyclic process $A B C A$ , as shown in the figure. Its pressure at $A$ is $p_{0}$ . Choose the correct option(s) from the following.

For the given cyclic process $C A B$ as shown for a gas, the work done is

In the given P-V diagram, a monoatomic gas $(γ = \frac{5}{3})$ is first compressed adiabatically from state A to state B. Then it expands isothermally from state B to state C. [Given: $(\frac{1}{3})^{0.6} ≃ 0.5, ln 2 ≃ 0.7$ ]. Which of the following statement(s) is(are) correct?

One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V-T) diagram. The correct statement(s) is/are: [R is the gas constant]