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:

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

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

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3.7904 J {m o l}^{- 1}

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

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Solution

The correct option is D $37.904 J {m o l}^{- 1}$
$H_{2} O (l) ⇌ H_{2} O (g)$
${Δ n}_{g} = 1$
$∴ Δ H = Δ V + Δ n_{g} R T$
$∴ Δ U = Δ H - Δ n_{g} R T$
$= 41 - 1 \times 8.314 \times 373 \times 10^{- 3}$
$∴ Δ U = 37.904 J / m o l$

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Similar questions

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

Δ U

) when

1 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}

)

Assuming that, water vapour is an ideal gas the internal energy change (ΔU)(ΔU) when 11 mol of water is vaporized at 11 bar pressure and 100o100o C will be:

[given: molar enthalpy of vaporisation of water at

1

bar and

373

= 41

m o l^{- 1}

and

R = 8.3

K^{- 1}

m o l^{- 1}

]

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