JEE MAIN - Chemistry (2011 - No. 6)
The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a
volume of 10 dm3 to a volume of 100 dm3 at 27oC is :
35.8 J mol-1 K−1
32.3 J mol-1 K−1
42.3 J mol-1 K−1
38.3 J mol-1 K−1
Explanation
Entropy change for an isothermal reversible process is given by
$$\Delta S = nR{\mkern 1mu} ln{{{V_2}} \over {{V_1}}}$$
$$ = 2 \times 8.314 \times 2.303log\,{{100} \over {10}}$$
$$ = 38.3\,J\,mo{l^{ - 1}}\,\,{K^{ - 1}}$$
$$\Delta S = nR{\mkern 1mu} ln{{{V_2}} \over {{V_1}}}$$
$$ = 2 \times 8.314 \times 2.303log\,{{100} \over {10}}$$
$$ = 38.3\,J\,mo{l^{ - 1}}\,\,{K^{ - 1}}$$
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