Consider the reversible isothermal expansion | vedclass

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(B) For a reversible isothermal expansion,the work done by the gas is given by $w = -nRT \ln(\frac{V_f}{V_i})$.Taking the magnitude,we have $|w| = nRT

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(B) For a reversible isothermal expansion,the work done by the gas is given by $w = -nRT \ln(\frac{V_f}{V_i})$.Taking the magnitude,we have $|w| = nRT \ln V_f - nRT \ln V_i$.This equation is in the form of a straight line $y = mx + c$,where $y = |w|$,$x = ln V_f$,slope $m = nRT$,and intercept $c = -nRT \ln V_i$.Since $T_2 > T_1$,the slope for $T_2$ $(nRT_2)$ is greater than the slope for $T_1$ $(nRT_1)$.Also,for a given expansion,the intercept $-nRT \ln V_i$ will be more negative for higher temperatures if we assume the same initial state or consistent expansion parameters,leading to the graph where the line for $T_2$ is steeper and has a lower intercept than the line for $T_1$.

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 $(|w|)$ on the final volume $(ln V)$ is:

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