Question

Mixing of $N_2$ and $H_2$ form an ideal gas mixture at room temperature in a container. For this process, which of the following statement is true?

• A. $\Delta H = 0,\ \Delta S_{surrounding}=0,\ \Delta S_{system}=0,\ \Delta G=-ve$
• B. $\Delta H = 0,\ \Delta S_{surrounding}=0,\ \Delta S_{system}>0,\ \Delta G=-ve$
• C. $\Delta H > 0,\ \Delta S_{surrounding}=0,\ \Delta S_{system}>0,\ \Delta G=-ve$
• D. $\Delta H < 0,\ \Delta S_{surrounding}>0,\ \Delta S_{system}<0,\ \Delta G=-ve$
• E. $\Delta H = 0,\ \Delta S_{surrounding}=0,\ \Delta S_{system}<0,\ \Delta G=-ve$

Hint:

Mixing of ideal gases always increases entropy.

Answer 1

The Correct Option is B

Concept:
Ideal gas mixing:
• No heat exchange $\Rightarrow \Delta H = 0$
• Entropy increases due to mixing

Step 1: Entropy change.
\[ \Delta S_{system} > 0 \]

Step 2: Free energy.
\[ \Delta G = \Delta H - T\Delta S < 0 \]