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What is the change in entropy of 1 mole of a perfect gas, which expands isothermally to ten times its original volume?

(a) 0.1 R
(b) 2.303 R
(c) 10.0 P
(d) 100.0 R
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  • $W_{rev} = -$2.303 nRT $log \frac{V_2}{V_1}$
  • Change in Entropy $\Delta S = \frac{q_{rev}}{T}$ = 2.303 nR $log \frac{V_2}{V_1}$
Answer: 2.303 R
We have, heat absorbed by the gas $q_{rev}= -W_{rev}$ = 2.303 nRT $log \frac{V_2}{V_1}$
Change in Entropy $\Delta S = \frac{q_{rev}}{T}$ = 2.303 nR $log \frac{V_2}{V_1}$
Here, n=1 and $\frac{V_2}{V_1} = 10$
$\therefore \Delta S = 2.303 \times 1 \times R \times log 10 = 2.303 \times R \times 1 =$ 2.303 R
answered Feb 26, 2014 by mosymeow_1

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