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:

$(d S)_{V, U} = 0, (d G)_{T, P}$ = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$(d S)_{V, U} = 0, (d G)_{T, P}$ = +ve

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$(d S)_{V, U} = - v e, (d G)_{T, P}$ = -ve

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$(d S)_{V, U} = + v e, (d G)_{T, P}$ = -ve

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$(d S)_{V, U} = + v e, (d G)_{T, P}$ = -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

Suggest Corrections

Similar questions

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

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)

and change in entropy

(Δ S)

satisfy the criteria:

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:

Q. Assertion :The thermodynamic factor which determines the spontaneity of a process is the free energy. For a process to be spontaneous the free energy must be -ve Reason: The change in free energy is related to the change in enthalpy and change in entropy. The change in entropy for a process must always be positive if it is spontaneous.

Q. Entropy change for an irreversible process taking system and surrounding together is :

Join BYJU'S Learning Program

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 (dS)V,U=+ve,(dG)T,P = -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
$(d S)_{V, U} = + v e, (d G)_{T, P}$ = -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