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- 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 beA- 41-00 kJ mol -1B- 4-100 kJ mol -1C- 4-100 kJ mol -1D- 4-100 kJ mol -1

The correct option is D 4.100 kJ mol^ 1 H = 41kJ mol^ 1 =41000 J mol^ 1 T = 100^∘ C = 273 + 100 = 373K n = 1 U = H nRT = 41000 (2 × 8.314 × 373) =37898 ...

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Standard XII

Chemistry

First Law of Thermodynamics

Assuming that...

Question

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

41.00 kJ mol $^{- 1}$

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4.100 kJ mol $^{- 1}$

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4.100 kJ mol $^{- 1}$

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4.100 kJ mol $^{- 1}$

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Solution

The correct option is D
4.100 kJ mol $^{- 1}$

$△ H = 41 k J m o l^{- 1} = 41000 J m o l^{- 1}$

$T = 100^{\circ} C = 273 + 100 = 373 K$

n = 1

$△ U = △ H - △ n R T = 41000 - (2 \times 8.314 \times 373)$

$= 37898.88 J m o l^{- 1} = 37.9 k J m o l^{- 1}$

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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}

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

(Δ U

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1

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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}

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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 (ΔU)(ΔU) when 11 mol of water is vaporized at 11 bar pressure and 100o100o C will be:

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373

= 41

m o l^{- 1}

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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∘ 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

The correct option is D 4.100 kJ mol−1 △H=41kJ mol−1 =41000Jmol−1 T=100∘C=273+100=373K n = 1 △U=△H−△nRT=41000−(2×8.314×373) =37898.88J mol−1=37.9 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

The correct option is D
4.100 kJ mol $^{- 1}$

$△ H = 41 k J m o l^{- 1} = 41000 J m o l^{- 1}$

$T = 100^{\circ} C = 273 + 100 = 373 K$

n = 1

$△ U = △ H - △ n R T = 41000 - (2 \times 8.314 \times 373)$

$= 37898.88 J m o l^{- 1} = 37.9 k J m o l^{- 1}$