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Standard XII

Chemistry

Heat of Formation

Standard enth...

Question

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:

43.76

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40.66

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37.56

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-43.76

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Solution

The correct option is D 37.56
$H_{2} O (l) ⟶ H_{2} O (g)$

$Δ n g = 1$

$Δ U = Δ H - Δ n g R T$

$= 40.66 - 1 \times 8.314 \times 373 \times 10^{- 3}$

$= 40.66 - 3.101$

$\approx + 37.56 K J / m o l$

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Similar questions

Q. Standard enthalpy of vapourisation

Δ_{v a p} H^{0}

for water at

100^{0} C

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:

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 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

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 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 $R = \frac{25}{3} J/molK$ )

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Standard enthalpy of vapourisation ΔvapH(0) for water at 100 is 40.66 kJmol−1. The internal energy of vapourisation of water at 100 (in kJmol−1) is:

The correct option is D 37.56H2O(l)⟶H2O(g)Δng=1ΔU=ΔH−ΔngRT=40.66−1×8.314×373×10−3=40.66−3.101≈+37.56KJ/mol

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:

The correct option is D 37.56
$H_{2} O (l) ⟶ H_{2} O (g)$

$Δ n g = 1$

$Δ U = Δ H - Δ n g R T$

$= 40.66 - 1 \times 8.314 \times 373 \times 10^{- 3}$

$= 40.66 - 3.101$

$\approx + 37.56 K J / m o l$