The molar heat capacity (\(\mathrm{C_p}\)) of \(\mathrm{CD_2O}\) is 10 cals at 1000 K. The change in entropy associated with cooling 32 g of \(\mathrm{CD_2O}\) vapour from 1000 K to 100 K at constant pressure will be: (D = deuterium, atomic mass = 2u)
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- \(23.03\ \mathrm{cal\,deg^{-1}}\)
- \(-23.03\ \mathrm{cal\,deg^{-1}}\)
- \(2.303\ \mathrm{cal\,deg^{-1}}\)
- \(-2.303\ \mathrm{cal\,deg^{-1}}\)
The Correct Option is B
Solution and Explanation
At constant pressure \(\Delta S = nC_p \ln\frac{T_2}{T_1}\).
\(\Delta S = 1 \times 10 \times \ln\frac{100}{1000} = 10 \times (-2.303) = -23.03\ \mathrm{cal\,deg^{-1}}\).
Answer: (2) \(-23.03\ \mathrm{cal\,deg^{-1}}\)
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