(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 \

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

(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 $A$.

Class 11 Chemistry Thermodynamics Free energy and Work

Medium

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

A
$\Delta G_T = nRT \ln \frac{P_2}{P_1}$
B
$\Delta G_T = nRT \ln \frac{V_2}{V_1}$
C
$\Delta G_T = nRT \log \frac{P_1}{P_2}$
D
$\Delta G_T = nRT \log \frac{V_2}{V_1}$

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Medium

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Medium

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

Similar Questions

For the reaction occurring in the gaseous phase: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,which of the following conditions is correct for the reaction to be spontaneous at high temperatures?

For a reaction $A + B \rightarrow$ products,$\Delta H = -84.2 \ kJ$ and $\Delta S = -200 \ J \ K^{-1}$. Calculate the highest value of temperature (in $K$) so that the reaction will proceed in the forward direction.

For the reaction $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$,$\Delta H = -285.8 \ kJ \ mol^{-1}$,$\Delta S = -0.163 \ kJ \ mol^{-1} K^{-1}$. What is the value of free energy change at $27 \ ^\circ C$ for the reaction in $kJ \ mol^{-1}$?

For the vaporization of water at $1 \ atm$ pressure,the values of $\Delta H$ and $\Delta S$ are $40.63 \ kJ/mol$ and $108.8 \ J/K \cdot mol$ respectively. At what temperature $(K)$ will the change in Gibbs energy $(\Delta G)$ be zero (in $.4$)?

For a given reaction,$\Delta H = 35.5 \ kJ \ mol^{-1}$ and $\Delta S = 83.6 \ J \ K^{-1} \ mol^{-1}$. The reaction is spontaneous at (Assume that $\Delta H$ and $\Delta S$ do not vary with temperature.)

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