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

The work done ($$w$$) during a reversible isothermal expansion from an initial volume $$V_i$$ to a final volume $$V$$ is given by:

$$w = -n\text{RT}\ln\left(\frac{V}{V_i}\right) = -n\text{RT}\ln V + n\text{RT}\ln V_i$$

Taking the magnitude (absolute value) of the work done ($$|w|$$) to match standard graphical representations:

$$|w| = n\text{RT}\ln V - n\text{RT}\ln V_i$$

This matches the straight-line equation $$y = mx + c$$ where:

• Y-axis ($$y$$): $$|w|$$
• X-axis ($$x$$): $$\ln V$$
• Slope ($$m$$): $$n\text{RT}$$ (Positive slope)
• Intercept ($$c$$): $$-n\text{RT}\ln V_i$$ (Negative intercept on the y-axis, meaning it intersects the x-axis at $$\ln V = \ln V_i$$ where $$|w| = 0$$)

Temperature Dependence:

Since the slope is directly proportional to temperature ($$\text{Slope} \propto \text{T}$$) and we are given $$\text{T}_1 < \text{T}_2$$:

$$\text{Slope for T}_2 > \text{Slope for T}_1$$

Intercept is directly proportional to temperature ($$\text{Slope} \propto \text{T}$$) and we are given $$\text{T}_1 < \text{T}_2$$:
$$\text{C for T}_2 > \text{C for T}_1$$
$$\text{Intercept for T}_2$$ will be more negative than $$\text{Intercept for T}_1$$

$$\text{Intercept for T}_2$$ lies below $$\text{Intercept for T}_1$$ on y-axis