Assuming that water vapour is an ideal gas, the internal energy change ( $Δ U$ ) when $1 m o l$ of water is vaporised at $1$ bar pressure and $100^{o} C$ will be:
(Given: Molar enthalpy of vapourisation of water at $1$ bar and $373 K = 41 k J . {m o l}^{- 1}$ and $R = 8.3 J {m o l}^{- 1}$ $K^{- 1}$ )

4.100 k J

{m o l}^{- 1}

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3.7904 k J

{m o l}^{- 1}

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37.904 k J

{m o l}^{- 1}

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41.00 k J

{m o l}^{- 1}

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Solution

The correct option is C $37.904 k J$ ${m o l}^{- 1}$
The change :
$H_{2} O (l) \to H_{2} O (g)$

$Δ H = Δ U + Δ n g R T$
"or" $Δ U = Δ H - Δ n g R T$

Substituting the values, we get
$Δ U = 41.00 k J / m o l - 1 \times 8.3 J / m o l / K \times 373 K \times \frac{1}{1000}$
$= 41.00 k J m o l^{- 1} - 3.096 k J m o l^{- 1}$
$= 37.904 k J / m o l$

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Similar questions

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

(Δ U

) when

1

mol of water is vaporised at

1

bar pressure and

100^{o} C

, (given: molar enthalpy of vaporisation of water at

1

bar and

373 K = 41 k J {m o l}^{- 1}

and

R = 8.314 J K^{- 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