Assuming that water vapour is an ideal gas, the internal energy ($\Delta \mathrm{U}$) when 1 mol of water is vapourised at 1 bar pressure and $100^{\circ} \mathrm{C}$, (Given: Molar enthalpy of vapourization of water at 1 bar and $373 \mathrm{~K}=41 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$) will be

(C) $37.904 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(A) $4.100 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(B) $3.7904 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(D) $41.00 \mathrm{~kJ} \mathrm{~mol}^{-1}$

JEE Main · 2007 · 29 Apr 2007 · Offline · Chemical Thermodynamics

Assuming that water vapour is an ideal gas, the internal energy (

Δ U

) when 1 mol of water is vapourised at 1 bar pressure and

10 0^{\circ} C

, (Given: Molar enthalpy of vapourization of water at 1 bar and

373 K = 41 kJ mol^{- 1}

and

R = 8.3 J mol^{- 1} K^{- 1}

) will be

Answer: (C)

For the vaporisation of one mole of water the change in the number of gas moles is

Δ n_{g} = 1

. Using

Δ H = Δ U + Δ n_{g} R T

, the internal-energy change is

Δ U = Δ H - Δ n_{g} R T = 41 - (8.3 \times 1 0^{- 3} \times 373)

kJ per mole. The correction term is about

3.1

kJ, so

Δ U \approx 37.9

kJ per mole.

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option