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Enthalpy of Solution

100 mL of a l...

Question

100mL of a liquid is contained in an insulated container at a pressure of 1 bar. The pressure is steeply increased to 100 bar. The volume of the liquid is decreased by 1mL at this constant pressure. The $Δ H$ and $Δ U$ are:

Δ H = 1 b a r m L

Δ U = 9900 b a r m L

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Δ H = 100 b a r m L

Δ U = 990 b a r m L

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Δ H = 100 b a r m L

Δ U = 9900 b a r m L

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Δ H = 1 b a r m L

Δ U = 9900 b a r m L

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Solution

The correct option is C $Δ H = 100 b a r m L$ $Δ U = 9900 b a r m L$
$Δ U = q + w$
For adiabatic process, $q = 0$ , hence $Δ U = w$
$Δ U = - 100 (99 - 100) = - 100 (- 1) = 100 b a r m L$
$Δ H = Δ U + Δ (p V)$
$= Δ U + (p_{2} V_{2} - p_{1} V_{1})$
$= 100 + (100$ x $99 - 1$ x $100$ )
$= 9900 b a r m L$

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100mL of a liquid is contained in an insulated container at a pressure of 1 bar. The pressure is steeply increased to 100 bar. The volume of the liquid is decreased by 1mL at this constant pressure. The ΔH and ΔU are:

The correct option is C ΔH= 100 bar mL ΔU= 9900 bar mLΔU=q+w For adiabatic process, q=0, hence ΔU=w ΔU=−100(99−100)=−100(−1)=100 bar mL ΔH=ΔU+Δ(pV) =ΔU+(p2V2−p1V1) =100+(100x99−1x100) =9900 bar mL

100mL of a liquid is contained in an insulated container at a pressure of 1 bar. The pressure is steeply increased to 100 bar. The volume of the liquid is decreased by 1mL at this constant pressure. The $Δ H$ and $Δ U$ are: