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Question

(A) For an isothermal process of an ideal gas,the change in Gibbs free energy is related to pressure by the equation: $\Delta G = \int_{P_1}^{P_2} V \

Vedclass · Accepted Answer

$\Delta G_T = nRT \ln \frac{P_2}{P_1}$

Vedclass · Answer

(A) For an isothermal process of an ideal gas,the change in Gibbs free energy is related to pressure by the equation: $\Delta G = \int_{P_1}^{P_2} V \, dP$. Since $PV = nRT$,we have $V = \frac{nRT}{P}$. Substituting this into the integral: $\Delta G = \int_{P_1}^{P_2} \frac{nRT}{P} \, dP = nRT \int_{P_1}^{P_2} \frac{1}{P} \, dP$. Evaluating the integral gives: $\Delta G = nRT \ln \left( \frac{P_2}{P_1} \right)$. Therefore,the correct option is $A$.

The dependence of Gibbs free energy on pressure for an isothermal process of an ideal gas is given by:

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