(B) For any spontaneous (irreversible) process in an isolated system,the change in entropy is positive,i.e.,$(dS)_{V,E} > 0$.For a process occurring a

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$(dS)_{V,E} > 0, (dG)_{T,P} < 0$

(B) For any spontaneous (irreversible) process in an isolated system,the change in entropy is positive,i.e.,$(dS)_{V,E} > 0$.For a process occurring at constant temperature $(T)$ and pressure $(P)$ involving only pressure-volume work,the spontaneity criterion is given by the change in Gibbs free energy,which must be negative,i.e.,$(dG)_{T,P} Therefore,the correct criteria are $(dS)_{V,E} > 0$ and $(dG)_{T,P} < 0$.

Class 11 Chemistry Thermodynamics Mix Examples-Thermodynamics and Thermochemistry

Medium

In an irreversible process taking place at constant $T$ and $P$ and in which only pressure-volume work is being done,the change in Gibbs free energy $(dG)$ and change in entropy $(dS)$ satisfy the criteria:

AIEEE 2003 All AIEEE

A
$(dS)_{V,E} < 0, (dG)_{T,P} < 0$
B
$(dS)_{V,E} > 0, (dG)_{T,P} < 0$
C
$(dS)_{V,E} = 0, (dG)_{T,P} = 0$
D
$(dS)_{V,E} = 0, (dG)_{T,P} > 0$

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In an irreversible process taking place at constant $T$ and $P$ and in which only pressure-volume work is being done,the change in Gibbs free energy $(dG)$ and change in entropy $(dS)$ satisfy the criteria:

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At $27 \, ^oC$,one mole of an ideal gas is compressed isothermally and reversibly from a pressure of $2 \ atm$ to $10 \ atm$. The values of $\Delta E$ and $q$ are $(R = 2 \ cal \ K^{-1} \ mol^{-1})$:

At $25^{\circ} \text{C}$,$1 \text{ mole}$ of butane is combusted to form $CO_2$ and liquid $H_2O$. The work done is $...... \text{ L atm}$.

$5 \, mol$ of an ideal gas at $100 \, K$ are allowed to undergo reversible compression till its temperature becomes $200 \, K$. If $C_v = 28 \, J \, K^{-1} \, mol^{-1}$,calculate $\Delta U$ and $\Delta pV$ for this process. $(R = 8.0 \, J \, K^{-1} \, mol^{-1})$

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