For water $Δ_{v a p} H = 41 k J {mol}^{- 1}$ at $373 K$ and 1 bar pressure. Assuming that water vapour is an ideal gas the that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___ $k J {mol}^{- 1}$ .
[Use : $R = 8.3 J {mol}^{- 1} K^{- 1}$ ]

Open in App

Solution

$H_{2} O (l) ⇌ H_{2} O (g) Δ H_{v a p} = 41 k J {mol}^{- 1}$

$Δ H = Δ E + Δ n_{g} R T$
$Δ E is internal energy change Δ n_{g} is change in gaseous moles = 1 - 0 = 1$

$41 = Δ E + 1 \times 8.3 \times 10^{- 3} \times 373$

$Δ E \approx (41 - 3) = 38 k J {mol}^{- 1}$

Suggest Corrections

Similar questions

Q. Assuming that water vapour is an ideal gas, the internal energy change

(Δ U)

when

1

mol of water is vaporized at

1

bar pressure and

100^{o}

C (given: molar enthalpy of vaporization of water

41

k J

m o l^{- 1}

1

bar and

373

K and

R = 8.3

m o l^{- 1}

m o l^{- 1}

) will be:

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