For water at 100^{circ} C and 1 , bar,Delta_{ | vedclass

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(D) The vaporization process is represented as: $H_{2}O_{(\ell)} \rightleftharpoons H_{2}O_{(g)}$From the first law of thermodynamics,the relation bet

Vedclass · Accepted Answer

$31$

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(D) The vaporization process is represented as: $H_{2}O_{(\ell)} ightleftharpoons H_{2}O_{(g)}$From the first law of thermodynamics,the relation between enthalpy change and internal energy change is given by: $\Delta H = \Delta U + \Delta n_{g} RT$Rearranging the equation: $\Delta H - \Delta U = \Delta n_{g} RT$For $1 \, mol$ of water vaporization,the change in the number of moles of gas is: $\Delta n_{g} = n_{g(product)} - n_{g(reactant)} = 1 - 0 = 1$Given $T = 100^{\circ} C = 373 \, K$ and $R = 8.31 \, J \, mol^{-1} \, K^{-1}$Substituting the values: $\Delta H - \Delta U = 1 \, mol 	imes 8.31 \, J \, mol^{-1} \, K^{-1} 	imes 373 \, K$$\Delta H - \Delta U = 3099.63 \, J \, mol^{-1}$Expressing in terms of $10^{2} \, J \, mol^{-1}$: $3099.63 \, J \, mol^{-1} \approx 31 	imes 10^{2} \, J \, mol^{-1}$Rounding to the nearest integer,the value is $31$.

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