If water vapour is assumed to be a perfect gas,the molar enthalpy change for the vaporisation of $1 \ mol$ of water at $1 \ bar$ and $100^{\circ} C$ is $41 \ kJ \ mol^{-1}$. Calculate the internal energy change when $1 \ mol$ of water is vaporised at $1 \ bar$ pressure and $100^{\circ} C$.
(N/A) The process is $H_2O_{(l)} \rightarrow H_2O_{(g)}$.
The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_g RT$.
Rearranging for internal energy change: $\Delta U = \Delta H - \Delta n_g RT$.
Here,$\Delta H = 41 \ kJ \ mol^{-1}$,$R = 8.314 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1}$,$T = 373 \ K$,and $\Delta n_g = 1 - 0 = 1$.
Substituting the values:
$\Delta U = 41 \ kJ \ mol^{-1} - (1 \times 8.314 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1} \times 373 \ K)$
$\Delta U = 41 \ kJ \ mol^{-1} - 3.101 \ kJ \ mol^{-1}$
$\Delta U = 37.899 \ kJ \ mol^{-1} \approx 37.9 \ kJ \ mol^{-1}$.
The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_g RT$.
Rearranging for internal energy change: $\Delta U = \Delta H - \Delta n_g RT$.
Here,$\Delta H = 41 \ kJ \ mol^{-1}$,$R = 8.314 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1}$,$T = 373 \ K$,and $\Delta n_g = 1 - 0 = 1$.
Substituting the values:
$\Delta U = 41 \ kJ \ mol^{-1} - (1 \times 8.314 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1} \times 373 \ K)$
$\Delta U = 41 \ kJ \ mol^{-1} - 3.101 \ kJ \ mol^{-1}$
$\Delta U = 37.899 \ kJ \ mol^{-1} \approx 37.9 \ kJ \ mol^{-1}$.
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DifficultNEET 2019
View SolutionThe enthalpy change on freezing of $1 \ mol$ of water at $5 \ ^\circ C$ to ice at $-5 \ ^\circ C$ is ..... $kJ \ mol^{-1}$.
(Given $\Delta _{fus}H = 6 \ kJ \ mol^{-1}$ at $0 \ ^\circ C$,
$C_p(H_2O, l) = 75.3 \ J \ mol^{-1} \ K^{-1}$,
$C_p(H_2O, s) = 36.8 \ J \ mol^{-1} \ K^{-1}$)
(Given $\Delta _{fus}H = 6 \ kJ \ mol^{-1}$ at $0 \ ^\circ C$,
$C_p(H_2O, l) = 75.3 \ J \ mol^{-1} \ K^{-1}$,
$C_p(H_2O, s) = 36.8 \ J \ mol^{-1} \ K^{-1}$)
DifficultJEE MAIN 2017
View SolutionCalculate $\Delta H^{\circ}$ for the reaction,$Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$,given the following reactions:
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2}H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2}H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$
DifficultAP EAMCET 2009
View SolutionCalculate the enthalpy change on freezing of $1.0 \ mol$ of water at $10.0^{\circ} C$ to ice at $-10.0^{\circ} C$. Given: $\Delta_{fus} H = 6.03 \ kJ \ mol^{-1}$ at $0^{\circ} C$,$C_p [H_2 O_{(l)}] = 75.3 \ J \ mol^{-1} \ K^{-1}$,$C_p [H_2 O_{(s)}] = 36.8 \ J \ mol^{-1} \ K^{-1}$.
Medium
View SolutionConsider the following data for the reaction $X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)$ at $600 \ K$. The $\Delta_r G^\circ$ (in $kJ \ mol^{-1}$) for the reaction is:
Compound
$\Delta_f H^\circ$ $(kJ \ mol^{-1})$
$S^\circ$ $(J \ mol^{-1} \ K^{-1})$
$XY(g)$
$42$
$200$
$X_2(g)$
$8$
$140$
$Y_2(g)$
$80$
$250$
| Compound | $\Delta_f H^\circ$ $(kJ \ mol^{-1})$ | $S^\circ$ $(J \ mol^{-1} \ K^{-1})$ |
|---|---|---|
| $XY(g)$ | $42$ | $200$ |
| $X_2(g)$ | $8$ | $140$ |
| $Y_2(g)$ | $80$ | $250$ |
DifficultJEE MAIN 2026
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