JEE MAIN - Chemistry (2021 - 26th August Evening Shift - No. 15)
For water $$\Delta$$vap H = 41 kJ mol$$-$$1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___________ kJ mol$$-$$1
[Use : R = 8.3 J mol$$-$$1 K$$-$$1]
[Use : R = 8.3 J mol$$-$$1 K$$-$$1]
Answer
38
Explanation
H2O(l) $$\to$$ H2O(g) : $$\Delta$$H = 41 $${{kJ} \over {mol}}$$
$$\Rightarrow$$ From the relation : $$\Delta$$H = $$\Delta$$U + $$\Delta$$ngRT
$$\Rightarrow$$ 41$${{kJ} \over {mol}}$$ = $$\Delta$$U + (1) $$\times$$ $${{8.3} \over {1000}}$$ $$\times$$ 373
$$\Delta$$ DU = 41 $$-$$ 3.0959 = 38 kJ/mol
$$\Rightarrow$$ From the relation : $$\Delta$$H = $$\Delta$$U + $$\Delta$$ngRT
$$\Rightarrow$$ 41$${{kJ} \over {mol}}$$ = $$\Delta$$U + (1) $$\times$$ $${{8.3} \over {1000}}$$ $$\times$$ 373
$$\Delta$$ DU = 41 $$-$$ 3.0959 = 38 kJ/mol
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