JEE MAIN - Chemistry (2021 - 27th July Morning Shift - No. 20)
For water at 100$$^\circ$$ C and 1 bar,
$$\Delta$$vap H $$-$$ $$\Delta$$vap U = _____________ $$\times$$ 102 J mol$$-$$1. (Round off to the Nearest Integer)
[Use : R = 8.31 J mol$$-$$1 K$$-$$1]
[Assume volume of H2O(l) is much smaller than volume of H2O(g). Assume H2O(g) treated as an ideal gas]
$$\Delta$$vap H $$-$$ $$\Delta$$vap U = _____________ $$\times$$ 102 J mol$$-$$1. (Round off to the Nearest Integer)
[Use : R = 8.31 J mol$$-$$1 K$$-$$1]
[Assume volume of H2O(l) is much smaller than volume of H2O(g). Assume H2O(g) treated as an ideal gas]
Answer
31
Explanation
H2O(l) $$\rightleftharpoons$$ H2O(v)
$$\Delta$$H = $$\Delta$$U + $$\Delta$$ngRT
For 1 mole waters;
$$\Delta$$ng = 1
$$\therefore$$ $$\Delta$$ngRT = 1 mol $$\times$$ 8.31 J/mol-k $$\times$$ 373 K
= 3099.63 J $$ \cong $$ 31 $$\times$$ 102 J
$$\Delta$$H = $$\Delta$$U + $$\Delta$$ngRT
For 1 mole waters;
$$\Delta$$ng = 1
$$\therefore$$ $$\Delta$$ngRT = 1 mol $$\times$$ 8.31 J/mol-k $$\times$$ 373 K
= 3099.63 J $$ \cong $$ 31 $$\times$$ 102 J
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