Note Chemistry, Physical Chemistry 1-19
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2 moles of an ideal gas, starting from the same initial conditions, namely V1=20 dm3 and p1=250 kPa, are compressed via (1) adiabatic reversible compression to p2=500 kPa； ; J c/ i( _! v% k- _ (2) Adiabatically, against a constant external pressure, with p(boundary)=500 kPa, compressed to equilibrium. Then the answer for ΔU1 (/ ) ΔU2 is
Using a graphical approach, you draw a P-V diagram. Since it’s an adiabatic process, dU = Q + W; therefore, dU = W. (1) dU1 = –fPdV, which represents the area under the arc. (2) dU2 = –P(constant) × (V2 – V1), which is the area of a rectangle. Compare dU1 and dU2
The book says, “When a system does work on its environment, the amount of work done in a reversible process is the maximum.”； When the environment does work on the system, the amount of work done in a reversible process is the smallest.” This case involves compression; since it is the environment that does work on the system, reversible compression results in the least amount of work, and consequently the increase in thermal energy is also the smallest