The Delta H for vaporisation of a liquid is 2 | vedclass

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(B) Given,$\Delta H_{vap} = 20 \, kJ/mol$. For the process of vaporisation,the change in the number of moles of gas is $\Delta n_g = n_{gas} - n_{liqu

Vedclass · Accepted Answer

$17.2$

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(B) Given,$\Delta H_{vap} = 20 \, kJ/mol$. For the process of vaporisation,the change in the number of moles of gas is $\Delta n_g = n_{gas} - n_{liquid} \approx 1 - 0 = 1$. Temperature $T = 60 + 273 = 333 \, K$. The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$. Rearranging for internal energy change: $\Delta U = \Delta H - \Delta n_g RT$. Substituting the values: $\Delta U = 20 \, kJ/mol - (1 \, mol 	imes 8.314 	imes 10^{-3} \, kJ \cdot K^{-1} \cdot mol^{-1} 	imes 333 \, K)$. $\Delta U = 20 - 2.768 \approx 17.23 \, kJ/mol$. Thus,the value is close to $17.2 \, kJ/mol$.

List-$I$ Thermodynamic Process	List-$II$ Magnitude in $kJ$
$A$. Work done in reversible,isothermal expansion of $2 \ mol$ of ideal gas from $2 \ dm^3$ to $20 \ dm^3$ at $300 \ K$.	$I$. $4$
$B$. Work done in irreversible isothermal expansion of $1 \ mol$ ideal gas from $1 \ m^3$ to $3 \ m^3$ at $300 \ K$ against a constant pressure of $3 \ kPa$.	$II$. $11.5$
$C$. Change in internal energy for adiabatic expansion of a $1 \ mol$ ideal gas with change of temperature $= 320 \ K$ and $\overline{C}_V = \frac{3}{2} R$.	$III$. $6$
$D$. Change in enthalpy at constant pressure of $1 \ mole$ ideal gas with change of temperature $= 337 \ K$ and $\overline{C}_P = \frac{5}{2} R$.	$IV$. $7$

The $\Delta H$ for vaporisation of a liquid is $20 \, kJ/mol$. Assuming ideal behaviour,the change in internal energy for the vaporisation of $1 \, mole$ of the liquid at $60^{\circ} C$ and $1 \, bar$ is close to $.... \, kJ/mol$

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