$1 \mathrm{gm}$ of water at a pressure of $1.01 \times 10^{5} \mathrm{~Pa}$ is converted into steam without…

$1 \mathrm{gm}$ of water at a pressure of $1.01 \times 10^{5} \mathrm{~Pa}$ is converted into steam without any change of temperature. The volume of $1 \mathrm{~g}$ of steam is 1671 $\mathrm{cc}$ and the latent heat of evaporation is $540 \mathrm{cal}$. The change in internal energy due to evaporation of $1 \mathrm{gm}$ of water is
  1. $\approx 167 \mathrm{cal}$
  2. $\approx 500 \mathrm{cal}$
  3. $540 \mathrm{cal}$
  4. $581 \mathrm{cal}$

Solution

$\begin{aligned} \mathrm{d} \mathrm{W} &=\mathrm{P} \Delta \mathrm{V}=1.01 \times 10^{5}[1671-1] \times 10^{-6} \mathrm{Joule} \\=& \frac{1.01 \times 167}{4.2} \mathrm{cal}=40 \mathrm{cal} \text { nearly } \\ \Delta \mathrm{Q}=\mathrm{mL}=1 \times 540, \\ \Delta \mathrm{Q}=\Delta \mathrm{W}+\Delta \mathrm{U} \\ \text { or } \Delta \mathrm{U}=540-40=500 \mathrm{cal} . \end{aligned}$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

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