Consider the reversible isothermal expansion of an ideal gas in a closed system at two different tem

Q: Why do we take the magnitude of work done?

The question explicitly plots $|W|$ on the y-axis. In expansion, work done by the gas is negative (by IUPAC convention), so taking the magnitude makes it positive for easier graphical representation in the first quadrant.

Q: Why is the slope dependent on temperature?

When we map the work equation to $y = mx + c$, the slope $m$ corresponds to $nRT$. Since $n$ and $R$ are constants for a given closed system, a higher temperature $T$ directly results in a steeper slope.

Q: Why do both lines in the correct graph start from the same point on the x-axis?

The x-intercept occurs where $|W| = 0$. Solving our equation gives $\ln V = \ln V_i$. Since both processes start from the same initial state, they share the same initial volume $V_i$, meaning they must intersect the x-axis at the exact same point.

Q: Can the y-intercept be positive?

Yes! The y-intercept is $-nRT \ln V_i$. If the initial volume $V_i$ is less than 1 (in the given units), $\ln V_i$ is negative, making the entire intercept positive.

Q: Why are graphs (a) and (b) incorrect if they show $T_2$ steeper than $T_1$?

While they get the slopes right, they fail the intercept test. Graph (a) shows the same y-intercept for different temperatures, which is mathematically impossible since $c = -nRT \ln V_i$. Graph (b) shows intercepts with different signs, which is also impossible for a fixed $V_i$.

Question

Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temp. $T_1$ and $T_2$ ($T_1 < T_2$). The correct graphical depiction of the dependence of work done ($W$) on the final volume ($V$) is

Sigma Percentile · Accepted Answer

### Decoding the Thermodynamics Graph: Work vs Volume

Graphs in thermodynamics aren't just arbitrary lines drawn on a grid; they are visual stories of physical processes. When you are asked to identify the correct graph for a process, you are essentially being asked to translate a physical law into the language of coordinate geometry. Let's break down this problem step-by-step and see how the math perfectly dictates the shape of the graph.

### The Master Equation

We are dealing with the reversible isothermal expansion of an ideal gas. The fundamental equation for the work done in this process is:

$$W = -nRT \ln\left(\frac{V_f}{V_i}\right)$$

The question specifically asks us to plot the **magnitude** of the work done, $|W|$, against the natural logarithm of the final volume, $\ln V$. Let's take the absolute value to make our expression positive (since $V_f > V_i$ in expansion, the logarithm is positive, and the negative sign makes $W$ negative by IUPAC convention):

$$|W| = nRT \ln\left(\frac{V}{V_i}\right)$$

Using the fundamental properties of logarithms, we can expand this fraction into a subtraction:

$$|W| = nRT \ln V - nRT \ln V_i$$

### The Geometry of Thermodynamics

Now, look closely at our expanded equation. It has the exact same structural DNA as the classic equation of a straight line:

$$y = mx + c$$

By mapping our variables, we get:
*   **y-axis ($y$):** $|W|$
*   **x-axis ($x$):** $\ln V$
*   **Slope ($m$):** $nRT$
*   **y-intercept ($c$):** $-nRT \ln V_i$

This mapping is our golden key. It tells us that the graph must be a straight line. But we have two different temperatures, $T_1$ and $T_2$, where $T_1 < T_2$. How does this affect the lines?

**1. Analyzing the Slope:**
The slope $m$ is directly proportional to the temperature $T$. Since $T_2 > T_1$, the slope for the $T_2$ process must be greater than the slope for the $T_1$ process ($m_2 > m_1$). Visually, this means the line for $T_2$ will be steeper. All four given options show $T_2$ as steeper, so we need to dig deeper.

**2. Analyzing the Intercepts:**
Let's find the **x-intercept**. The x-intercept is the point where the line crosses the x-axis, which means the y-value ($|W|$) is zero. Setting $|W| = 0$ in our equation:

$$0 = nRT \ln V - nRT \ln V_i$$

Dividing both sides by $nRT$ (which is non-zero), we get:

$$\ln V = \ln V_i$$

This is a profound result! The x-intercept is simply $\ln V_i$. Notice that **temperature ($T$) has completely vanished from this expression**.

### The Final Verdict

Because both the $T_1$ and $T_2$ processes are expansions of the *same* gas starting from the *same* closed system setup, they must share the exact same initial volume, $V_i$.

Consequently, both lines must intersect the x-axis at the exact same point: $\ln V_i$.

If we look at the given options:
*   **Graph (a)** shows the lines starting from the same point on the y-axis, which implies they have the same y-intercept. But $c = -nRT \ln V_i$, so different temperatures mean different y-intercepts. This is incorrect.
*   **Graph (b)** and **Graph (d)** show the lines having y-intercepts with different signs or one being zero. Since $V_i$ is constant, the term $\ln V_i$ has a fixed sign. Therefore, both y-intercepts must share the same sign. These are incorrect.
*   **Graph (c)** shows both lines originating from the exact same point on the positive x-axis and diverging with $T_2$ being steeper. This perfectly aligns with our mathematical deduction that they share a common x-intercept.

Thus, Graph (c) is the undisputed correct depiction of the process.

Select Answer:

Visualized Solution

Graphical Analysis

Work Done in Isothermal Expansion

Magnitude and Expansion

Equation of a Straight Line

Comparing Slopes

Finding the X-Intercept

Analyzing Y-Intercepts

Conclusion

The Sigma Insight: First Law of Thermodynamics

Decoding the Thermodynamics Graph

The Master Equation

The Geometry of Thermodynamics

The Final Verdict

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