MCQ
For a reversible spontaneous change $\Delta S$ is
- A$\frac{{\Delta E}}{T}$
- B$\frac{{P\,\Delta V}}{T}$
- ✓$\frac{q}{T}$
- D$RT\, log\, K$
✓
Answer
Correct option: C.
$\frac{q}{T}$
c
$ds =\frac{ dq \cdot rev }{ T }$
$ds =\frac{ dq \cdot rev }{ T }$
$\Delta s =\frac{\int dq \cdot rev }{ T }$
$\int dq \cdot$ rev.q (for rev. provess)
$\Delta s =\frac{ q }{ T }$
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