(a) Initial state = A = (1.00 atm, 5.00 L, 300K), so by n = p*V/(R*T), n = (1 atm)*(5 L)/(0.0821 L*atm/((mol*K)*(300 K)) = 0.203 mol
Warmed isochorically to a pressure of 3 atm, so
T = p*V/(n*R) = (3.00 atm)*(5.00 L)/((0.203 mol)*(0.0821 L*atm/(mol*K))) = 900 K, so B = (3.00 atm, 5 L, 900 K)
Expanded isothermally to 1.00 atm, so, obviously, the temperature remains constant. What is the volume at point C?
V = n*R*T/p = ((0.203 mol)*(0.0821 L*atm/(mol*K)))/(1 atm) = 15.0 L, so C = (1.00 atm, 15.0 L, 900 K)
And finally compressed isobarically back to state A.
To calculate the work done in each of these processes, you need to recognize that if w is the work done *on* the system, δw = -p*dV
Because A -> B is an isochoric (constant volume) process, dV = 0, so w = 0
For B -> C, we know that the temperature is constant, then p = n*R*T/V so δw = -n*R*T/V dV. This can be integrated to get w = -n*R*T*ln(V_final/V_initial) = (0.203 mol)*(0.0821 L*atm/(mol*K))*(900 K)*ln(15/3) = -2,446 J
For C -> A, because this is an isobaric compression, w = -p*ΔV = -(1 atm)*(-10 L) = 1,013 J
It's easiest to next look at the internal energy changes.
Cv for a monatomic ideal gas is 3R/2.
For A -> B, an isochoric process, ΔE = n*Cv*ΔT = (0.203 mol)*(3R/2)*(600K) = (0.203 mol)*(3/2)(8.314 J/(mol*K))*(600K) = 1,519 J [your answer is correct, though you are reporting too many sig figs, given the sig figs of the data, you should report this as 1,520 J]
B -> C is an isothermal process. The internal energy of an ideal gas depends only on the temperature, so if the temperature is constant, ΔE = 0.
Finally, we compress at constant pressure back to the original state, but the internal energy is a function of state, so the internal energy has to come back to the same value it had initially. The change in internal energy for C -> A must be equal to the negative of the energy change from A -> B (because ΔE = 0 for B -> C), so ΔE = -1,520 J for C -> A.
Now use the first law to calculate q. If q is the thermal energy added *to* the system in a process, ΔE = = q + w, q = ΔE - w, and:
For A -> B, q = 1,519 J - 0 J= 1,519 J
For B -> C, q = 0 - (-2,446 J) = 2,446 J
For C -> A, q = -1,519 J - 1,013 J = -2,532 J
For the whole cyclic process, we've already seen that ΔE = 0. Adding up the heat added to the system, the net thermal ennergy transferred to the system in the cycle is: (1,519 + 2,446 - 2,532)J = 1,433 J and the work done on the system is: (0 - 2,446 + 1,013) J = -1,433 J. which makes sense because ΔE = 1,433 J + (-1,433 J) = 0 for the entire process.