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Standard XII

Chemistry

Internal Energy

The molar ent...

Question

The molar enthalpy change for $H_{2} O (1) ⇌ H 2 O (g)$ at $373$ K and $1$ atm is $41$ kJ/mol. Assuming ideal behavior, the internal energy change for vaporization of $1$ mol of water at $373$ K and $1$ atm in kJ $m o l^{- 1}$ is $?$

30.2

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41.0

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48.1

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37.9

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Solution

The correct option is D $37.9$

$W = - n R T = 1 (8.314 \times 373) \times 10^{- 3} = - 3.10 K J$
$q = 41 K J$
$Δ U = 37.9 K J$

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Similar questions

Q. The molar enthalpy change for

H_{2} O (l) ⇌ H_{2} O (g)

373

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^{- 1}

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

(Δ U)

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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

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The molar enthalpy change for H2O(1)⇌H2O(g) at 373K and 1 atm is 41kJ/mol. Assuming ideal behavior, the internal energy change for vaporization of 1 mol of water at 373K and 1 atm in kJ mol−1 is ?

The correct option is D 37.9W=−nRT=1(8.314×373)×10−3=−3.10KJq=41KJΔU=37.9KJ

The molar enthalpy change for $H_{2} O (1) ⇌ H 2 O (g)$ at $373$ K and $1$ atm is $41$ kJ/mol. Assuming ideal behavior, the internal energy change for vaporization of $1$ mol of water at $373$ K and $1$ atm in kJ $m o l^{- 1}$ is $?$

The correct option is D $37.9$

$W = - n R T = 1 (8.314 \times 373) \times 10^{- 3} = - 3.10 K J$
$q = 41 K J$
$Δ U = 37.9 K J$