Question

When unit mass of water boils to become steam at $100C^{\circ }$ , it absorbs Q amount of heat. The densities of water and steam at $100C^{\circ }$ are $p_1$ and $p_2$ respectively and the atmospheric pressure is $p_0$. The increase in internal energy of water is

• A. $Q$
• B. $Q+p_0(\frac{1}{p_1}-\frac{1}{p_2})$
• C. $Q+p_0(\frac{1}{p_2}-\frac{1}{p_1})$
• D. $Q-p_0(\frac{1}{p_1}-\frac{1}{p_2})$

Answer 1

The correct option is B $Q+p_0(\frac{1}{p_1}-\frac{1}{p_2})$

Work done $dU=dQ-dW$

Work is done against atmospheric pressure which is given as $P_0$.

Since the densities of water and steam at ${100}^{0}C$ are given as $p_1$ and $p_2$ respectively,

$dW=P_0(V_2-V_1)=P_0(\frac{m}{p_2}-\frac{m}{p_1})=P_0\times 1(\frac{1}{p_2}-\frac{1}{p_1})$ since $m=1$ as given.

$\therefore dU=Q-P_0\times 1(\frac{1}{p_2}-\frac{1}{p_1})=Q+P_0(\frac{1}{p_1}-\frac{1}{p_2})$

Hence,option B is correct.