(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 + \

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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 $R = 8.314 \times 10^{-3}\ kJ\ K^{-1}\ mol^{-1}$,we have:$40.66 = \Delta U + (1 \times 8.314 \times 10^{-3} \times 373)$.$40.66 = \Delta U + 3.101$.$\Delta U = 40.66 - 3.101 = 37.559\ kJ\ mol^{-1} \approx 37.56\ kJ\ mol^{-1}$.

Class 11 Chemistry Thermodynamics First law of thermodynamics

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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)

AIPMT 2012 All AIPMT

A
$+ 37.56$
B
$-43.76$
C
$+43.76$
D
$+ 40.66$

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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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At constant $T$ and $P$,which one of the following statements is correct for the reaction,$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}$?

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What is the difference between $\Delta H$ and $\Delta U$ for the reaction given below at $298 \ K$ (in $kJ$)? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$
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