# Unit mass of a liquid with volume V₁ is completely changed into a gas of volume V₂ 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(V₂-V₁) C L - P(V₂-V₁) D L

## 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 $L$

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## Answer

The correct option is **C**: $$L - P(V_{2} - V_{1})$$

To arrive at this, let's recall the first law of thermodynamics, which states:

$$ \Delta Q = \Delta U + P \Delta V $$

In this problem, $\Delta Q$ is given as the latent heat of evaporation $L$ for unit mass. The external pressure is constant and is denoted by $P$. The volumes before and after the phase change are $V_1$ and $V_2$, respectively.

Given the unit mass (i.e., $m = 1$), the latent heat $L$ can be directly used:

$$ L = \Delta U + P(V_{2} - V_{1}) $$

From this, we can isolate $\Delta U$:

$$ \Delta U = L - P(V_{2} - V_{1}) $$

Thus, the final increase in the internal energy of the system is:

$$ \boxed{L - P(V_{2} - V_{1})} $$

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