Question

Unit mass of liquid of volume $V_1$ completely turns into a gas volume $V_2$ at constant atmospheric pressure $P$ and temperature $T$. The latent heat of vaporization is ${}^{\prime }{L}^{\prime }$. The change in internal energy of the gas is:

• A. $L$
• B. $L+P(V_2-V_1)$
• C. $L-P(V_2-V_1)$
• D. zero

Answer 1

Answer: (C)

$L-P(V_2-V_1)$

When the heat energy is supplied to the liquid, it is absorbed by the liquid and a phase change occurs from liquid state to gaseous state. The amount of heat absorbed by the liquid molecules is transferred to the vapour molecules.
Thus latent heat 'L' is given by,

$L=\frac{Q}{m}$

where Q is the amount of heat energy supplied to the liquid and m is the mass of the liquid.
For a unit mass of liquid,

$L=Q$

The latent heat of vaporization is the amount of heat energy supplied by heating, according to first law of thermodynamics,
$L=\delta E+P\delta V$
$\delta E=L-P\delta V$
$\delta E=L-P(V_2-V_1)$
where,
$\delta E$ is a change in internal energy of the gas,
P is atmospheric pressure,
$\delta V$ is a change in volume
V1 is the volume of liquid and
V2 is the volume of gas

Correct option is 'C'