Question

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

• A. 41.00 kJ mol${}^{-1}$
• B. 4.100 kJ mol${}^{-1}$
• C. 4.100 kJ mol${}^{-1}$
• D. 4.100 kJ mol${}^{-1}$

Answer 1

The correct option is D

4.100 kJ mol${}^{-1}$

$△H=41kJmo{l}^{-1}=41000Jmo{l}^{-1}$

$T={100}^{\circ }C=273+100=373K$

n = 1

$△U=△H-△nRT=41000-(2\times 8.314\times 373)$

$=37898.88Jmo{l}^{-1}=37.9kJmo{l}^{-1}$