Standard enthalpy of vapourisation $Δ_{v a p} H^{0}$ for water at $100^{0} C$ is $40.66 k J . m o l^{- 1}$ . The internal energy change of vapourisation of water at $100^{0} C$ $(i n k J . m o l^{- 1})$ is:

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Solution

We know that,
For gaseous reactant and products we have a relation between standard enthalpy of vaporization $(Δ_{v a p} H^{0})$ and internal energy $(Δ V)$

$Δ_{V a p} H^{0} = Δ V^{0} + Δ n_{g} R T$ .

Where, $Δ_{n g} = n_{2} - n_{1}$ , different between number of moles of reactant and product.

For vaporization of water:
$H_{2} O (l) \to H_{2} O (g)$

So, change in no. of moles $= Δ_{n g} = 1 - 0 = 1$
Thus,
$\begin{matrix} Δ_{v a p} H^{0} = Δ_{v a p} U^{0} + R T Δ_{v a p} U^{0} = Δ_{v a p} H^{0} - R T = 40.66 - (8.314 \times 10^{- 3} \times 373 K) = 40.66 - 3.10 k J / m o l = 37.55 k J / m o l \end{matrix}$

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Q. Standard enthalpy of vapourisation

Δ_{v a p} H^{(0)}

for water at

100

is 40.66

k J {m o l}^{- 1}

. The internal energy of vapourisation of water at

100

(in

k J {m o l}^{- 1}

) is:

If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourization of 1 mol of water at 1 bar and

100^{0} C

41 k J . {m o l}^{- 1}

. Calculate the internal energy change, when 1 mol of water is vaporized at 1 bar pressure and

100^{0} C

Assuming that water vapour is an ideal gas, the internal energy change $(Δ U)$ when 1 mol of water is vapourised at 1 bar pressure and 100 $^{0}$ C will be:

[Given molar enthalpy of vapourisation of water at 1bar and 373 K=41 kJ mol $^{- 1}$ ]

Assuming the water vapour is an ideal gas, the internal energy change $(Δ U)$ in kJ when 1 mole of water is vapourised at $177^{\circ} C$ . (Given molar enthalpy of vapourisation of water $177^{\circ} C$ = 37 kJ/mol and $R = \frac{25}{3} J/molK$ )

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

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Standard enthalpy of vapourisation ΔvapH0 for water at 1000C is 40.66 kJ.mol−1. The internal energy change of vapourisation of water at 1000C (in kJ.mol−1) is:

Standard enthalpy of vapourisation $Δ_{v a p} H^{0}$ for water at $100^{0} C$ is $40.66 k J . m o l^{- 1}$ . The internal energy change of vapourisation of water at $100^{0} C$ $(i n k J . m o l^{- 1})$ is: