Question

Assuming ideal behaviour, calculate the work done when 1.6 mole of water evaporates at 373 K against the atmospheric pressure of 1 atm.

• A. 29.67 J
• B. 48.8 J
• C. 21.2 J
• D. none of these

Answer 1

The correct option is D none of these

Solution:- (D) none of these

Volume of $1.6$ mole of liquid water $\left(V_1\right)$-

As we know that,

$\text{Density}=\frac{\text{mass}}{\text{volume}}$

$\therefore V_1=\frac{1.6\times 18}{1}=28.8mL=0.0288L[\text{Density of water}=1gm/mL]$

Using ideal gas equation,

$PV=nRT\Rightarrow V=\frac{nRT}{P}$

Volume of $1.6$ mole of water at $373K$ in gaseous state $\left(V_2\right)$-

$V_2=\frac{nRT}{P}=\frac{1.6\times 0.0821\times 373}{1}=48.93L$

Now as we know that,

$W=-P_{ext.}\Delta V$

Given that $P_{ext.}=1atm$

$\therefore W=-1(48.93-0.0288)=-4954.8J$