Consider 1 ,kg of liquid water undergoing a c | vedclass

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Question

(B) The change in volume $\Delta V$ for $1 \,kg$ of water during phase change is given by:$\Delta V = V_{	ext{vapour}} - V_{	ext{liquid}} = \frac{m}

Vedclass · Accepted Answer

$20.8 	imes 10^5$

Vedclass · Answer

(B) The change in volume $\Delta V$ for $1 \,kg$ of water during phase change is given by:$\Delta V = V_{	ext{vapour}} - V_{	ext{liquid}} = \frac{m}{ho_{	ext{vapour}}} - \frac{m}{ho_{	ext{liquid}}}$$\Delta V = \frac{1}{(1/1.8)} - \frac{1}{1000} = 1.8 - 0.001 \approx 1.8 \,m^3$Work done by the system against constant pressure $p$ is:$W = p \Delta V = (1.01 	imes 10^5 \,N m^{-2}) 	imes (1.8 \,m^3) = 1.818 	imes 10^5 \,J$Heat absorbed during phase change is:$Q = mL = 1 \,kg 	imes 22.6 	imes 10^5 \,J kg^{-1} = 22.6 	imes 10^5 \,J$Using the first law of thermodynamics,$\Delta U = Q - W$:$\Delta U = 22.6 	imes 10^5 - 1.818 	imes 10^5$$\Delta U = 20.782 	imes 10^5 \,J kg^{-1} \approx 20.8 	imes 10^5 \,J kg^{-1}$.

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