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There is 1 mo! liquid (molar volume 100 mL) in an adiabatic container initial, pressure being I bar. Now the pressure is steeply increased to 100 bar, and the volume decreased by 1 mL under constant pressure of 100 bar. Calculate $Δ H$ and $Δ E$ . [Given 1 bar $= 10^{5} N / m^{2}$ ]

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$Δ E = 20 J, Δ H = 890 J$

$Δ E = 0 J, Δ H = 10 J$

$Δ H = 990 J, Δ E = 10 J$

$Δ E = 0 J, Δ H \neq 0 J$

answer is B.

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Detailed Solution

Here,

$q = 0$ (adiabatic)

$\begin{array}{l} Δ U = w = - 100 \times \frac{- 1}{1000} = 0.1 L-bar \\ Δ H = Δ U + Δ (P V) = 0.1 + \frac{(100 \times 99 - 1 \times 100)}{1000} \\ = 9.9 L-bar \end{array}$

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