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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

Assuming that water vapor is an ideal gas, the internal energy change ($ \Delta U$) when 1 mole of water is vaporized at 1 bar pressure and $100^{\circ}C$ (molar enthalpy of vaporization of water at 1 bar and 373 K = 41 kJ/mol and R = 8.314 J/Kmol) will be

$\begin {array} {1 1} (A)\;4.100\: kJ/mol & \quad (B)\;3.7904\: kJ/mol \\ (C)\;37.904\: kJ/mol & \quad (D)\;41.00\: kJ/mol \end {array}$

1 Answer

$H_2O(l) \rightarrow H_2O(s)$
$ \Delta n = 1-0 = 1$
$ \Delta E = \Delta H - \Delta nRT$
$= 41 - 1 \times 8.3 \times 373 \times10^{-3} = 37.9\: kJ/mol$
Ans : (C)
answered Mar 18, 2014 by thanvigandhi_1
 

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