Given that $\Delta H = 0$ for the mixing of two gases,explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) For the mixing of two gases,the enthalpy change is $\Delta H = 0$.
According to the Gibbs free energy equation,$\Delta G = \Delta H - T\Delta S$.
Since $\Delta H = 0$,the equation becomes $\Delta G = -T\Delta S$.
When two gases mix,the entropy of the system increases,meaning $\Delta S > 0$.
Therefore,$\Delta G$ becomes negative $(\Delta G < 0)$,which indicates that the process is spontaneous.

Explore More

Similar Questions

What is the entropy change in $J K^{-1}$ during the melting of $27.3 \ g$ of ice at $0^{\circ} C$? (Latent heat of fusion of ice $= 330 \ J g^{-1}$)

Which of the following processes is associated with a decrease in entropy?

Which of the following is the correct expression for an irreversible process?

Standard entropy of $X_2, Y_2$ and $XY_3$ are $60, 40$ and $50 \ J \ K^{-1} \ mol^{-1},$ respectively. For the reaction,$\frac{1}{2}X_2 + \frac{3}{2}Y_2 \to XY_3, \ \Delta H = -30 \ kJ,$ to be at equilibrium,the temperature will be ............... $K$.

Given the standard entropy values at $298 \ K$ and $1 \ atm$ as $H_2(g) : 130.6 \ J \ K^{-1} \ mol^{-1}$,$Cl_2(g) : 223.0 \ J \ K^{-1} \ mol^{-1}$,and $HCl(g) : 186.7 \ J \ K^{-1} \ mol^{-1}$,calculate the entropy change $(\Delta S^{\circ})$ in $J \ K^{-1} \ mol^{-1}$ for the reaction: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo