A vessel contains 100 litres of a liquid $x$ . Heat is supplied to the liquid in such a fashion that, Heat given = change in enthalpy. The volume of the liquid increases by 2 litres. If the external pressure is one atm, and 202.6 Joules of heat were supplied then, [ U - total internal energy ]

$Δ U = 0, Δ H = 0$

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$Δ U = + 202.6 J, Δ H = + 202.6 J$

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$Δ U = - 202.6 J, Δ H = - 202.6 J$

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$Δ U = 0, Δ H = + 202.6 J$

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Solution

The correct option is D
$Δ U = 0, Δ H = + 202.6 J$

$Δ H$ = heat gives so process is isobaric.

$w = - P_{n e x t} (Δ V) = - 1 \times 2 l i t r e . a t m = - 202.6 J$ -as one litre.atm=101.33 joules

Hence $q = 202.6 J = Δ H$ and $Δ E = q + w = 0$

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