One mole of a liquid ( $1 b a r, 100 m l$ ) is taken in an adiabatic container and the pressure increases steeply to $100 b a r$ . Then at constant pressure of $100 b a r$ , volume decrases by $1 m L$ . Calculate $Δ H$ .

500 J

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990 J

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650 J

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720 J

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Solution

The correct option is B $990 J$
From first law of thermodynamics,
$Δ U = q + w$
But as the process is carried out under adiabatic condition, $q = 0$ .
Hence $Δ U = w$
Further, at constant pressure of $100 b a r$ , volume has decreased by $1 m L$ , therefore, work of contraction will be :
$w = P Δ V = 100 b a r \times 1 m L$
$= (100 \times 10^{5} N m^{- 2}) (10^{- 6} m^{3}) = 10 J$

Hence, $Δ U = 10 J$
Enthapy change is given as:
$Δ H = Δ U + Δ (P V)$
$Δ (P V) = (P_{2} V_{2} - P_{1} V_{1})$

$P_{1} = 1 b a r, V_{1} = 100 m L$
$P_{2} = 100 b a r, V_{2} = 99 m L$

$Δ (P V) = (100 \times 99) - (1 \times 100)$
$Δ (P V) = 9800 b a r m L = 980 J$
as $1 L b a r = 101.3 J$
so,
$Δ H = 10 J + 980 J = 990 J$

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One mole of a liquid (1 bar, 100 ml) is taken in an adiabatic container and the pressure increases steeply to 100 bar. Then at constant pressure of 100 bar, volume decrases by 1 mL. Calculate ΔH.

One mole of a liquid ( $1 b a r, 100 m l$ ) is taken in an adiabatic container and the pressure increases steeply to $100 b a r$ . Then at constant pressure of $100 b a r$ , volume decrases by $1 m L$ . Calculate $Δ H$ .