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

= 0

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= +ve

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= -ve

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= -ve

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Solution

The correct option is D
= -ve

$Δ G = Δ H - T Δ S$
Since T is constant, $D e l t a H = 0$
Also, as the process is spontaneous, we can conclude from the above equation that:
$△$ G = -ve
$△$ S = +ve

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

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Q. 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

(Δ G)

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A gas is expanded from volume $V_{0} t o 2 V_{0}$ under three different processes, as shown in the figure. Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic. Let $Δ U_{1}, Δ U_{2}$ and $Δ U_{3}$ be the change in the internal energy of the gas in these three processes, then:

One mole of diatomic gas is taken through a cyclic process. Pressure, volume and temperature of gas at A is $P_{0}, V_{0}$ and $T_{0}$ . List-I gives some physical quantities and List-II gives their possible magnitudes.

List-I

List-II

(1) Work done in process BC	(P) $\frac{3}{2} P_{0} V_{0}$
(2) Work done in process CA	(Q) $2 P_{0} V_{0}$
(3) Change in internal energy during process BC	(R) $3 P_{0} V_{0}$
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	(U) 0

If cyclic process is given on P-T graph as shown in figure. Then match List-I with List-II.

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

The correct option is D = -ve ΔG=ΔH−TΔS Since T is constant, DeltaH=0 Also, as the process is spontaneous, we can conclude from the above equation that: △G = -ve △S = +ve

The correct option is D
= -ve

$Δ G = Δ H - T Δ S$
Since T is constant, $D e l t a H = 0$
Also, as the process is spontaneous, we can conclude from the above equation that:
$△$ G = -ve
$△$ S = +ve