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The work done on the system when one mole of an ideal gas at 500 K is compressed isothermally and reversibly to 1/10th of its original volume (R = 2 cal)

(a) 500 kcal
(b) 1.51 kcal
(c) -23.03 kcal
(d) 2.303 kcal
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  • Work done on the system W= - 2.303 nRT log$\frac{V_2}{V_1}$
Answer: 2.303 kcal
Since, work done on the system W= - 2.303 nRT log$\frac{V_2}{V_1}$
W = $-2.303\times 1 \times 2 \times 500$ log $\frac{1}{10} = \frac{-2.303\times 1000 log 10^{-1}}{1000}$ kJ = + 2.303 kJ
answered Feb 24, 2014 by mosymeow_1
 

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