test 1 question

I have a question about this problem -- a monatomic gas in a 3.5 liter container is originally at 27 C and atmospheric pressure. It is heated at constant volume until its temperature is 180 C, then at constant pressure until the gas has increased its volume by .52 liters. How much thermal energy is required? By how much does the internal energy of the gas change? How much work is done in the process?

In the first part of the problem (volume is constant) Vo=3.5L, To=300K, Po=1atm and we know that temp changes to 453K. The first law of thermodynamics tells us that when V is contant W=0 and Q='dU. Since this is a monatomic gas I used 'dU=nCv'dt. One mol occupies 22.4L at stp so 3.5L is about .156mol. 'dU=.156mol(12.465J/molK)(153K)=297.5J = Q.

For the second part of the problem P is constant so Q='dU+W=P'dV, and volume increases by .52L. I am not sure where to go from here, or if what I did in the first part of this problem is correct.

The first part looks great.
You know `dV. From the first part you can find the pressure P. So you can find P `dV, which is the work done by the system in the expansion.
You can also use the ideal gas law to find the temperature required to expand the gas to its new volume. The expansion is at constant pressure, so you can find the energy required by using Cp = 5/2 R.
The change in internal energy is just the energy required to increase the KE of the particles. The gas is monatomic so it has only three degrees of freedom, so the change in internal energy is just 3/2 R `dT.
Having found the work, the energy put into the system and the change in the internal energy of the system, you can check that Q really does equal `dU + `dW.