JEE MAIN - Chemistry (2019 - 10th April Evening Slot - No. 7)
The difference between $$\Delta $$H and $$\Delta $$U ($$\Delta $$H – $$\Delta $$U), when the combustion of one mole of heptane(l) is carried
out at a temperature T, is equal to :
– 4 RT
3 RT
– 3 RT
4 RT
Explanation
We know,
$$\Delta $$H - $$\Delta $$U = $$\Delta $$ngRT
C7H16($$l$$) + 11O2(g) $$ \to $$ 7CO2(g) + 8H2O($$l$$)
Here $$\Delta $$ng = 7 - 11 = - 4
$$ \therefore $$ $$\Delta $$H - $$\Delta $$U = - 4RT
$$\Delta $$H - $$\Delta $$U = $$\Delta $$ngRT
C7H16($$l$$) + 11O2(g) $$ \to $$ 7CO2(g) + 8H2O($$l$$)
Here $$\Delta $$ng = 7 - 11 = - 4
$$ \therefore $$ $$\Delta $$H - $$\Delta $$U = - 4RT
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