Question

When $1gm$ of water changes from liquid to vapour phase at constant pressure of $1$ atmosphere, the volume increases from $1cc$ to $1671cc$. The heat of vaporisation at this pressure is $540cal/gm$. Increase in internal energy of water is
$(1atmosphere=1.01\times {10}^{6}dyne/c{m}^2)$.

• A. $4200J$
• B. $8200J$
• C. $1200J$
• D. $2100J$

Answer 1

The correct option is D $2100J$

Change in Internal energy = Q in - work done

It s a constant pressure work

$\therefore wD=P(V_2-V_1)$

$=1.01\times {10}^5[1.671-1]\times {10}^{-6}$

$=169.21s$

$P=1atm$

$=\mathrm{1.0.1}\times {10}^{5}Pa$

$=1.01\times {10}^6\frac{dyne}{c{m}^2}$

Latent heat of vaporization $=540\frac{cal}{gm}\times \left(1gm\right)$

$=540cal$

$1cal=4.184J.$

$=540\times 4.184$

$=2259.36J$

$\therefore \Delta U=2259.36-169.21$

$=2090Joules$

$\approx 2100Joules$