[mrusnak]

The problem is...
A monoatomic ideal gas is expanded reversibly at constant pressure of 4.00 atm, 10.0 L
and 300 K to 20.0 L. It is then cooled at constant volume to 300 K. Calculate q and w.

My question is...
Since dw=-pdV, how do you account for the temperature change when calculating work done? Do you solve the number of moles of the gas using P, V(initial), and T(initial), then solve for T(final)? Or does the temperature change not matter because it is cooled at constant volume back to the initial temperature?

[Yggdrasil]

The temperature change doesn't affect the work done.  In the expansion step, dw = -pdV and in the second step w = 0 because there is no change in volume.  The change in temperature will have an effect on the internal energy of the gas because  :delta: U = nC_v :delta: T