$A$ swimmer coming out from a pool is covered with a film of water weighing about $18 \ g$. How much heat must be supplied to evaporate this water at $298 \ K$? Calculate the internal energy of vaporisation at $298 \ K$. Given: $\Delta_{vap} H^{\ominus}$ for water at $298 \ K = 44.01 \ kJ \ mol^{-1}$.
(N/A) The process of evaporation is represented as: $H_2O_{(l)} \rightarrow H_2O_{(g)}$.
Number of moles in $18 \ g$ of $H_2O$ is $n = \frac{18 \ g}{18 \ g \ mol^{-1}} = 1 \ mol$.
Heat supplied to evaporate $18 \ g$ of water at $298 \ K$ is $q = n \times \Delta_{vap} H^{\ominus} = 1 \ mol \times 44.01 \ kJ \ mol^{-1} = 44.01 \ kJ$.
To calculate the internal energy of vaporisation $(\Delta_{vap} U)$,we use the relation: $\Delta_{vap} U = \Delta_{vap} H^{\ominus} - \Delta n_g RT$.
Here,$\Delta n_g = 1$ (since $1 \ mol$ of gas is produced from $1 \ mol$ of liquid).
$\Delta_{vap} U = 44.01 \ kJ - (1 \ mol) \times (8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) \times (298 \ K)$.
$\Delta_{vap} U = 44.01 \ kJ - 2.48 \ kJ = 41.53 \ kJ$.
Number of moles in $18 \ g$ of $H_2O$ is $n = \frac{18 \ g}{18 \ g \ mol^{-1}} = 1 \ mol$.
Heat supplied to evaporate $18 \ g$ of water at $298 \ K$ is $q = n \times \Delta_{vap} H^{\ominus} = 1 \ mol \times 44.01 \ kJ \ mol^{-1} = 44.01 \ kJ$.
To calculate the internal energy of vaporisation $(\Delta_{vap} U)$,we use the relation: $\Delta_{vap} U = \Delta_{vap} H^{\ominus} - \Delta n_g RT$.
Here,$\Delta n_g = 1$ (since $1 \ mol$ of gas is produced from $1 \ mol$ of liquid).
$\Delta_{vap} U = 44.01 \ kJ - (1 \ mol) \times (8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) \times (298 \ K)$.
$\Delta_{vap} U = 44.01 \ kJ - 2.48 \ kJ = 41.53 \ kJ$.
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Similar Questions
For the following reaction $Fe_2O_{3(s)} + 3 CO_{(g)} \longrightarrow 2 Fe_{(s)} + 3 CO_{2(g)}$,$\Delta H^{\circ} = -29.8 \ kJ$ and $\Delta S^{\circ} = 15 \ JK^{-1}$. What is the value of $\Delta S_{\text{total}}$ at $298 \ K$ (in $JK^{-1}$)?
Match the following columns:
Column $I$ Column $II$ $(a)$ Adiabatic process $(1)$ Heat $(b)$ Isolated system $(2)$ Constant volume $(c)$ Isothermal change $(3)$ First law of thermodynamics $(d)$ Path function $(4)$ No exchange of matter and energy $(e)$ State function $(5)$ No heat exchange $(f)$ $\Delta U = q$ $(6)$ Constant temperature $(g)$ Law of conservation of energy $(7)$ Internal energy $(h)$ Reversible process $(8)$ $p_{ext} = 0$ $(i)$ Free expansion $(9)$ Constant pressure $(j)$ $\Delta H = q$ $(10)$ Infinitely slow process involving equilibrium states $(k)$ Intensive property $(11)$ Entropy $(l)$ Extensive property $(12)$ Pressure $(13)$ Specific heat
| Column $I$ | Column $II$ |
| $(a)$ Adiabatic process | $(1)$ Heat |
| $(b)$ Isolated system | $(2)$ Constant volume |
| $(c)$ Isothermal change | $(3)$ First law of thermodynamics |
| $(d)$ Path function | $(4)$ No exchange of matter and energy |
| $(e)$ State function | $(5)$ No heat exchange |
| $(f)$ $\Delta U = q$ | $(6)$ Constant temperature |
| $(g)$ Law of conservation of energy | $(7)$ Internal energy |
| $(h)$ Reversible process | $(8)$ $p_{ext} = 0$ |
| $(i)$ Free expansion | $(9)$ Constant pressure |
| $(j)$ $\Delta H = q$ | $(10)$ Infinitely slow process involving equilibrium states |
| $(k)$ Intensive property | $(11)$ Entropy |
| $(l)$ Extensive property | $(12)$ Pressure |
| $(13)$ Specific heat |
Medium
View SolutionWhich of the following is/are true about the reversible isothermal expansion of an ideal gas?
$(a) \ \Delta U = 0$
$(b) \ q = 0$
$(c) \ \Delta T = 0$
$(d) \ q = 2.303 \ nRT \ \log_{10} \left( \frac{V_2}{V_1} \right)$
$(a) \ \Delta U = 0$
$(b) \ q = 0$
$(c) \ \Delta T = 0$
$(d) \ q = 2.303 \ nRT \ \log_{10} \left( \frac{V_2}{V_1} \right)$
Medium
View SolutionThree moles of an ideal gas are expanded isothermally from a volume of $300 \ cm^3$ to $2.5 \ L$ at $300 \ K$ against a constant external pressure of $1.9 \ atm$. The work done in joules is:
DifficultMHT CET 2019
View SolutionConsider the following statements:
Statement-$I$: During isothermal expansion of an ideal gas,its enthalpy decreases.
Statement-$II$: When $2.0 \ L$ of an ideal gas expands isothermally into vacuum,$\Delta U = 0$.
Statement-$I$: During isothermal expansion of an ideal gas,its enthalpy decreases.
Statement-$II$: When $2.0 \ L$ of an ideal gas expands isothermally into vacuum,$\Delta U = 0$.
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