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(C) The work done during the phase change at constant pressure is given by $W = P \Delta V$.Given $P = 1.013 	imes 10^5\,Pa$ (atmospheric pressure),$

Vedclass · Accepted Answer

$2089$

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(C) The work done during the phase change at constant pressure is given by $W = P \Delta V$.Given $P = 1.013 	imes 10^5\,Pa$ (atmospheric pressure),$V_{liquid} = 1\,cm^3 = 10^{-6}\,m^3$,and $V_{vapour} = 1671\,cm^3 = 1671 	imes 10^{-6}\,m^3$.$W = 1.013 	imes 10^5 	imes (1671 - 1) 	imes 10^{-6} \approx 169\,J$. Using the standard approximation $P = 10^5\,Pa$ as implied by the options: $W = 10^5 	imes 1670 	imes 10^{-6} = 167\,J$.The heat supplied is $Q = mL = 1\,g 	imes 2256\,J/g = 2256\,J$.From the first law of thermodynamics,$Q = \Delta U + W$,so $\Delta U = Q - W$.$\Delta U = 2256\,J - 167\,J = 2089\,J$.

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