Question
A gas expands adiabatically against a constant external pressure of 3.0 atm from 3.00 L to 5.00 L. Assuming only boundary work and negligible kinetic and potential energy changes, find its internal-energy change to the nearest joule. Use 1 atm = 101325 Pa.
The correct answer is
−608 J
Adiabatic expansion gives Q = 0. The work is \(W=p_{ext}\Delta V=3(101325)(0.002)=607.95\ \mathrm J\). Therefore \(\Delta U=Q-W=-607.95\ \mathrm J\approx-608\ \mathrm J\).
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