ThermodynamicsHard

Question

Heat liberated by a given amount of an ideal gas undergoing reversible isothermal process is 1200 cal at 300 K. What is the Gibbs free energy change of the gas in this process?

Options

A.Zero
B.+1200 cal
C.−1200 cal
D.4 cal

Solution

$q = \Delta U - w = 0 - \left( - nRT.\ln\frac{P_{1}}{P_{2}} \right) = nRT.\ln\frac{P_{1}}{P_{2}} = - \Delta G$

$\therefore\Delta G = - q = - ( - 1200) = + 1200\text{ cal}$

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