Question

The molar heat capacity $\left(C_p\right)$ of $C{D}_{2}O$ is $10$ cals at $1000K$. The change in entropy associated with cooling of $32g$ of $C{D}_{2}O$ vapour from from $1000K$ to $100K$ at constant pressure will be:
($D=$ deuterium, at. mass $=2u$)

• A. $-2.303calde{g}^{-1}$
• B. $23.03calde{g}^{-1}$
• C. $-23.03calde{g}^{-1}$
• D. $2.303cal{\mathrm{deg}}^{-1}$

Answer 1

Answer: (C)

Solution:- (B) $-23.03cal/deg$

$\Delta S=n{C}_p\ln \frac{T_2}{T_1}=2.303n{C}_p\log \frac{T_2}{T_1}$

Given:-

Mass of $C{D}_{2}O=32g$

Molecular mass of $C{D}_{2}O=32g$

$\therefore n=\frac{32}{32}=1\text{mole}$

$C_p=10cal$

$T_1=1000K$

$T_2=100K$

$\therefore \Delta S=2.303\times 1\times 10\times \log \frac{100}{1000}$

$\Rightarrow \Delta S=23.03(\log 1-\log 10)$

$\Rightarrow \Delta S=-23.03cal/deg(\because \log 1=0)$

Hence the change in entropy is $-23.03cal/deg$.