Standard enthalpy of vaporisation Delta_{vap} | vedclass

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(A) The process is $H_2O_{(l)} \longrightarrow H_2O_{(g)}$.The relation between enthalpy change and internal energy change is $\Delta H = \Delta U + \

Vedclass · Accepted Answer

$+ 37.56$

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(A) The process is $H_2O_{(l)} \longrightarrow H_2O_{(g)}$.The relation between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.Here,$\Delta H = 40.66\ kJ\ mol^{-1}$,$T = 100 + 273 = 373\ K$,and $\Delta n_g = 1 - 0 = 1$.Using $R = 8.314 	imes 10^{-3}\ kJ\ K^{-1}\ mol^{-1}$,we have:$40.66 = \Delta U + (1 	imes 8.314 	imes 10^{-3} 	imes 373)$.$40.66 = \Delta U + 3.101$.$\Delta U = 40.66 - 3.101 = 37.559\ kJ\ mol^{-1} \approx 37.56\ kJ\ mol^{-1}$.

Standard enthalpy of vaporisation $\Delta_{vap}H^o$ for water at $100\ ^oC$ is $40.66\ kJ\ mol^{-1}.$ The internal energy of vaporisation of water at $100\ ^oC$ (in $kJ\ mol^{-1}$) is ............... $kJ\ mol^{-1}$ (Assume water vapour to behave like an ideal gas)

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