There is 1 mol liquid (molar volume 100 ml) in an adiabatic container initial, pressure being 1 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 ]
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Δ H = 990 J, Δ E = 10 J
Δ U = q + W
for adiabatic process q = 0, hence Δ U = W and W = – p( Δ V) = – P (V 2 – V 1 )
so, Δ U = – 100 (99 – 100) = – 100 (–1) = 100 bar mL = 10 J
Now Δ H = Δ U + Δ (PV)
Here Δ U already calculated above and
Δ PV = (P 2 V 2 – P 1 V 1 )
So, Δ H = 100 + (100 × 99 – 1 × 100) = 9900 bar mL = 990 J
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