Why does adiabatic expansion do less PV work than Isothermal expansion?

[PeterPesto]

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I think it's because the temperature of the system is decreasing, less pressure is exerted, and thus less pressure volume work is done. Whereas in isothermal expansion, the PV = constant holds, according to Boyle's Law.

Is this correct?

 

[Rabolisk]

Yes. The exact derivation for adiabatic expansion (that is, a formula that relates P, V, and T) is beyond the scope of the MCAT, but temperature decreases.

You should be able to justify why temperature decreases in adiabatic expansion though (reversible or irreversible).

 

[PeterPesto]

Do you mind elaborating?

Temperature decreases because it is doing work on its surroundings?

Also, I'm assuming that temperature decreases more for a reversible expansion because it happens in a greater number of steps?

 

[eryaman]

In an ideal gas, internal energy is a function of ONLY the temperature of the gas.

dU = dW + dQ, where dU=change in internal energy of the gas, dW=work done on the gas, and dQ=heat added to the gas (the First Law of Thermodynamics)

In an isothermal expansion:
The gas expands, meaning it does work. Therefore, dW is negative. However, since the expansion is isothermal, the temperature stays constant and dU = 0 (since U(T) only). Therefore, dQ must greater than 0.

In order for isothermal expansion to occur, heat must be added from the surroundings.

In an adiabatic expansion:
Adiabatic means dQ = 0, therefore dU = dW. Because the gas expands, dW is negative and dU is negative as well. Because U is a function of only T and a decrease in U means a decrease in T, temperature decreases in an adiabatic expansion. This does not happen in an isothermal expansion because heat is added to the gas from the surroundings to keep the temperature constant.

Hope this helps!