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Standard XII

Chemistry

Enthalpy

Latent heat o...

Question

Latent heat of vapourisation of a liquid at $500 K$ and $1$ atm pressure is $10 k c a l / m o l$ . What is the change in internal energy when $3$ moles of the liquid is vapourised at the same temperature.

27 k c a l

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7 k c a l

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33 k c a l

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25 k c a l

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Solution

The correct option is A $27 k c a l$
$3 A (l) ⟶ 3 A (g)$

$△ n = 3 - 0 = 3$

$△ E = △ H - △ n R T$

$= 3 \times 10 - 3 \times 500 \times 0.002$

$= 27 k c a l$

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Latent heat of vapourisation of a liquid at 500K and 1 atm pressure is 10 kcal/mol. What is the change in internal energy when 3 moles of the liquid is vapourised at the same temperature.

The correct option is A 27kcal3A(l)⟶3A(g)△n=3−0=3△E=△H−△nRT=3×10−3×500×0.002=27kcal

Latent heat of vapourisation of a liquid at $500 K$ and $1$ atm pressure is $10 k c a l / m o l$ . What is the change in internal energy when $3$ moles of the liquid is vapourised at the same temperature.

The correct option is A $27 k c a l$
$3 A (l) ⟶ 3 A (g)$

$△ n = 3 - 0 = 3$

$△ E = △ H - △ n R T$

$= 3 \times 10 - 3 \times 500 \times 0.002$

$= 27 k c a l$