reversible expansion of 2 mole of an ideal gas
from a volume of 10dm3 to a volume of 100dm3
at 27°C is
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Answer: the answer is = 38.3 J mol-1K-1
Entropy for isothermal process -
\Delta S= nR\ln \frac{V_{f}}{V_{i}}
\Delta S= nR\ln \frac{P_{i}}{P_{f}}
- wherein
T_{f}=T_{i}
\Delta T=0
\Delta S= nRln \frac{V1}{V2}=2.303 nR \log \frac{V_{1}}{V_{2}}
=2.303 \times 2 \times 8.314 \times log \frac{100}{10}
= 38.3 J mol-1K-1
Explanation:
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