A sample consisting of 2.00 mole of perfect gas molecules, for which Cv,m = R, initially at P1=111 kPa and T1= 277 K, is heated reversibly to 356 K at constant volume. Calculate the final Pressure, Delta U, q, and w.

Given:
n = 2.00 moles
Cv,m = R
P1 = 111 kPa
T1 = 277 K
T2 = 356 K

First, we need to calculate the initial volume using the ideal gas law: PV = nRT

V1 = (nRT1) / P1
V1 = (2.00 mol * 8.314 J/mol K * 277 K) / (111000 Pa)
V1 = 0.413 m^3

Since the process is constant volume, we have W = 0

Now, we can calculate the final pressure using the ideal gas law with the constant volume condition: P1/T1 = P2/T2

P2 = P1 * (T2 / T1)
P2 = 111000 Pa * (356 K / 277 K)
P2 = 142447 Pa

Next, we can calculate the change in internal energy using the first law of thermodynamics:
ΔU = nCv,mΔT
ΔU = 2.00 mol * 8.314 J/mol K * (356 K - 277 K)
ΔU = 1345.6 J

Finally, we can calculate the heat added to the system:
q = ΔU + W
q = 1345.6 J + 0
q = 1345.6 J

Therefore, the final pressure is 142447 Pa, the change in internal energy is 1345.6 J, the heat added to the system is 1345.6 J, and the work done by the system is 0 J.