For a spontaneous process, free energy
A. Is zero
B. Increases
C. Decreases whereas the entropy increases
D. And entropy both decrease
Answer: Option C
Solution(By Examveda Team)
From second law of thermodynamics$$\eqalign{ & TdS \geqslant \delta Q \cr & \Rightarrow TdS \geqslant dU + \delta W \cr} $$
For an irreversible process $$TdS - dU - \delta W > 0$$
And for a reversible process $$TdS - dU - \delta W = 0$$
For any spontaneous process there should be finite changes so, we can consider it as an irreversible process and we know for irreversible process from second law of thermodynamics by above discussion: $$TdS - dU - PdV > 0$$
Under constant temperature and volume process $$ - dF > 0 \Rightarrow dF < 0$$
Similarly for an constant temperature and pressure process $$d\left( {TS - U - PV} \right) > 0 \Rightarrow dG < 0$$
Related Questions on Chemical Engineering Thermodynamics
A. Maxwell's equation
B. Thermodynamic equation of state
C. Equation of state
D. Redlich-Kwong equation of state
Henry's law is closely obeyed by a gas, when its __________ is extremely high.
A. Pressure
B. Solubility
C. Temperature
D. None of these
A. Enthalpy
B. Volume
C. Both A & B
D. Neither A nor B
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