Standard enthalpy of vapourisation _vapH^ for water at 100^ is 40.66 kJ/mol . The internal energy of vapourisation of water at 100^ (in kJ/mol ) is: - Chemistry MCQ Chemistry | ExamTest.live

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Standard enthalpy of vapourisation Δ_{vap} H^{\circ} for water at 10 0^{\circ} C is 40.66 kJ/mol . The internal energy of vapourisation of water at 10 0^{\circ} C (in kJ/mol) is:

A $+ 40.66$
B $+ 37.56$
C $- 43.76$
D $+ 43.76$

Solution & Step-by-step Explanation

For the process H_{2} O(l) \to H_{2} O(g) : Δ H = Δ U + Δ n_{g} RT Given: Δ H = 40.66 kJ/mol = 40660 J/mol Δ n_{g} = 1 - 0 = 1 R = 8.314 J/(mol K) T = 10 0^{\circ} C = 373 K $ Δ U = Δ H - Δ n_{g} RT $ Δ U = 40660 - (1 \times 8.314 \times 373) Δ U = 40660 - 3101.1 \approx 37558.9 J/mol Δ U \approx 37.56 kJ/mol

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Standard enthalpy of vapourisation Δ_{vap} H^{\circ} for water at 10 0^{\circ} C is 40.66 kJ/mol . The internal energy of vapourisation of water at 10 0^{\circ} C (in kJ/mol) is:

+ 40.66

+ 37.56

- 43.76

+ 43.76

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Standard enthalpy of vapourisation $Δ_{vap} H^{\circ}$ for water at $10 0^{\circ} C$ is $40.66 kJ/mol$ . The internal energy of vapourisation of water at $10 0^{\circ} C$ (in $kJ/mol$ ) is:

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Standard enthalpy of vapourisation Δvap​H∘ for water at 100∘C is 40.66kJ/mol. The internal energy of vapourisation of water at 100∘C (in kJ/mol) is:

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Standard enthalpy of vapourisation $Δ_{vap} H^{\circ}$ for water at $10 0^{\circ} C$ is $40.66 kJ/mol$ . The internal energy of vapourisation of water at $10 0^{\circ} C$ (in $kJ/mol$ ) is: