ThermodynamicsHard
Question
A liquid is adiabatically expanded from state – I to state – II, suddenly by a single step, as shown in the figure then
Options
A.$\Delta H = \frac{2\gamma P_{o}V_{o}}{\gamma - 1}$
B.$\Delta U = \frac{3P_{o}V_{o}}{\gamma - 1}$
C.ΔH = −PoVo
D.ΔU = −3PoVo
Solution
$q = 0 \Rightarrow \Delta U = w = - P_{0}\left( 4V_{0} - V_{0} \right) = - 3P_{0}V_{0}$
Now, $\Delta H = \Delta U + \Delta(PV) = \left( - 3P_{0}V_{0} \right) + \left( P_{0}.4V_{0} - 2P_{0}.V_{0} \right) = - P_{0}.V_{0}$
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