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.
$$P\left( {{V_2} - {V_1}} \right)$$
C.
$$L - P\left( {{V_2} - {V_1}} \right)$$
D.
$$L$$
Answer :
$$L - P\left( {{V_2} - {V_1}} \right)$$
Solution :
$$\eqalign{
& Q = mL = 1 \times L = L;W = P\left( {{V_2} - {V_1}} \right) \cr
& {\text{Now }}Q = \Delta U + W \cr
& {\text{or}}\,\,L = \Delta U + P\left( {{V_2} - {V_1}} \right) \cr
& \therefore \Delta U = L - P\left( {{V_2} - {V_1}} \right) \cr} $$
Releted MCQ Question on Heat and Thermodynamics >> Thermodynamics
Releted Question 1
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