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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The molar heat capacity for an ideal gas at constant pressure is $20.785 \ J \ K^{-1} \ mol^{-1}$. The change in internal energy is $5000 \ J$ upon heating it from $300 \ K$ to $500 \ K$. The number of moles of the gas is [Nearest integer] (Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)
Arrange the following in order of magnitude of work done by the system / on the system at constant temperature:
$(a)$ $|w_{reversible}|$ for expansion in infinite stage.
$(b)$ $|w_{irreversible}|$ for expansion in single stage.
$(c)$ $|w_{reversible}|$ for compression in infinite stage.
$(d)$ $|w_{irreversible}|$ for compression in single stage.
Choose the correct answer from the options given below:
$(a)$ $|w_{reversible}|$ for expansion in infinite stage.
$(b)$ $|w_{irreversible}|$ for expansion in single stage.
$(c)$ $|w_{reversible}|$ for compression in infinite stage.
$(d)$ $|w_{irreversible}|$ for compression in single stage.
Choose the correct answer from the options given below:
How much heat in $KJ$ is released during the combustion of $112 \ L$ of water gas at $S.T.P.$? (Water gas is an equimolar mixture of $H_2$ and $CO$)
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)}; \Delta H = -241.8 \ KJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -283 \ KJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)}; \Delta H = -241.8 \ KJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -283 \ KJ$
Easy
View SolutionEnthalpy of neutralisation of acetic acid by $NaOH$ is $-50.6 \ kJ/mol$ and the heat of neutralisation of a strong acid with a strong base is $-55.9 \ kJ/mol$. What is the value of $\Delta H$ for the ionisation of $CH_3COOH$ in $kJ/mol$?
DifficultAIIMS 2000
View SolutionThe enthalpies of combustion of carbon and carbon monoxide are $-393.5 \, kJ \, mol^{-1}$ and $-283 \, kJ \, mol^{-1}$ respectively. The enthalpy of formation of carbon monoxide per mole is $....... \, kJ \, mol^{-1}$. (in $.5$)
MediumAIEEE 2004
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