Question

Unit mass of a liquid with volume ${V_1}$​ is completely changed into a gas of volume ​${V_2}$ at a constant external pressure $P$ and temperature $T$ . If the latent heat of evaporation for the given mass is $L$ , then the increase in the internal energy of the system is?
A. Zero
B. $P({V_2} - {V_1})$
C. $L - P({V_2} - {V_1})$
D. $L$

Answer 1

Hint: Generally when external input energy is required to change the state from liquid into gases or vapor at constant temperature and constant external pressure is called the latent heat of vaporization. So, we can find out the increase internal energy of the system by using work done formula from the first law of thermodynamics, which is mentioned in below.

Formula used:
The formulae used in the solution of this problem are: -
$\Delta U = Q - W$
$Work Done = W = P\Delta V = P({V_2} - {V_1})$

Complete answer:
We know that the First Law of Thermodynamics is the application of conservation of energy and according to the first law: -
$\Delta U = Q - W$ … (1)
where $\Delta U = $ Change in Internal Energy
$Q = $ Heat Added to the system
$W = $ Work done by the system

Now, Pressure-Volume Work in thermodynamics is defined as: -
At constant pressure $Workdone = W = P\Delta V = P({V_2} - {V_1})$

From eq. (1), we get
$\Delta U = mL - P({V_2} - {V_1})$
$\left( {\therefore Q = m \times L} \right)$
where $L = $ Latent heat of evaporation (given)
and since the unit mass of liquid is taken, therefore $m = 1$

Thus, the increase in the internal energy of the system is: -
$ \Rightarrow \Delta U = (1)L - P({V_2} - {V_1}) = L - P({V_2} - {V_1})$

Hence, the correct option is (C) $L - P({V_2} - {V_1})$.

Note: Gases have higher internal energy as compared to solid and liquid state. This is because lot of energy is required to boind up the bonds between atoms or molecules and as a result this huge amount of energy provides a negative contribution to the internal energy in the solid and liquid state.