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Chemistry Question on Thermodynamics

Assuming that water vapour is an ideal gas, the internal energy (?U)(?U) when 1 mol of water is vapourised at 1 bar pressure and 100??100^??, (Given: Molar enthalpy of vapourization of water at 1 bar and 373 K=41 kJ mol−1373 \,K = 41\, kJ \,mol^{-1} and R=8.3 J mol−1K−1R = 8.3\, J \,mol^{-1}K^{-1}) will be

A

4.100 kJ mol−14.100\, kJ\, mol^{-1}

B

3.7904 kJ mol−13.7904\, kJ\, mol^{-1}

C

37.904 kJ mol−137.904\, kJ\, mol^{-1}

D

41.00 kJ mol−141.00\, kJ\, mol^{-1}

Answer

37.904 kJ mol−137.904\, kJ\, mol^{-1}

Explanation

Solution

H2O(ℓ)−>[vaporisation]H2O(g) {H_{2}O(\ell)->[vaporisation]H2O(g)} Δng=1−0=1\Delta n_{g}=1-0=1 ΔH=ΔU+ΔngRT\Delta H=\Delta U+\Delta n_{g}RT ΔU=ΔH−ΔngRT\Delta U=\Delta H-\Delta n_{g}RT =41−8.3×10−3×373=41-8.3\times10^{-3}\times373 =37.9 kJ mol−1=37.9\,kJ\,mol^{-1} Hence, (3) is correct.