Question

A heating coil is immersed in a 100 g sample of $H_{2}O\left(l\right)$ at 1 atm and ${100}^{\circ }C$ in a closed vessel. In this heating process, 60% of the liquid is converted in to gaseous form at constant pressure of 1 bar. Densities of liquid and gaseous water under these condition are 1000$kg/m^3,0.60kg/m^3$ respectively. magnitude of the work done for the process is :

• A. 4997 J
• B. 3970 J
• C. 9994 J
• D. None of these

Answer 1

The correct option is C 9994 J

Density $\left(d\right)$ = $\frac{mass}{volume}$
Initial volume of 100 g sample of $H_{2}O\left(V_i\right)$ = $\frac{Mass}{density}$
$=\frac{0.1kg}{1000kg{m}^{-3}}=1\times {10}^{-4}{m}^3$
Since the efficiency is 60%, hence 60 g of water is converted to gas and 40 g of water remains same.
Final volume $\left(V_f\right)$ = Volume of 60 g gas + volume of 40 g of water
$=V_f=\frac{0.06}{0.6}+\frac{0.04}{1000}$
So work done = $-p_{ext.}(V_f-V_i)$
$=-{10}^5(\frac{0.06}{0.60}+\frac{0.04}{1000}-1\times {10}^{-4})=-{10}^5(0.1+4\times {10}^{-5}-1\times {10}^{-4})|w|=9994J$