posted by AC on .
Suppose 2500 J of heat are added to 4.1 mol of argon gas at a constant pressure of 120 kPa. (Assume that the argon can be treated as an ideal monatomic gas.) Find the change in internal energy.
I started using U = N (3/2 kB T). But I have too many unknowns. The only other equation that our teacher gave us is Won + Qon = deltaU (Work on system + Heat on system = Change in internal energy). Please help!
Qon = deltaU - Won =
deltaU + Wsystem,
where Wsystem is the work performed by the system (in this case the argon gas). Because the pressure is constant, the work performed is:
P deltaV = Delta (PV) = (using ideal gas law) Delta (N kB T)
This means that:
Qon = deltaU + Wsystem =
5/2 N kB delta T =
(5/2)(2/3)(3/2)N kB delta T =
(5/2)(2/3) delta U = 5/3 delta U
And it follows that:
delta U = 3/5 Qon = 1500 J