Given:
- Number of moles of gas (n) = 3 mol
- Initial temperature (T1) = 200 K
- Initial pressure (P1) = 2 atm
- Final temperature (T2) = 250 K
- Molar heat capacity at constant volume (Cv) = 27.5 J/K·mol

To Find:
- Work done during the process

Assumptions:
- The gas follows the ideal gas law.
- The process is reversible and adiabatic.

Solution:

1. Calculation of Initial Volume (V1):
Using the ideal gas law, we can calculate the initial volume of the gas.
PV = nRT
V1 = (nRT1) / P1
= (3 mol * 0.0821 L·atm/(mol·K) * 200 K) / (2 atm)
= 24.84 L

2. Calculation of Final Volume (V2):
Since the process is adiabatic and reversible, we can use the relationship between temperature and volume for an adiabatic process.
T1 * V1^(γ-1) = T2 * V2^(γ-1)

The gas constant (R) can be expressed as R = Cp - Cv, where Cp is the molar heat capacity at constant pressure. For an ideal gas, γ = Cp / Cv.
Given that Cp = Cv + R, we can substitute the values to find γ.

Cp = Cv + R
Cp = 27.5 J/K·mol + 8.314 J/(K·mol)
Cp = 35.814 J/K·mol

γ = Cp / Cv
= 35.814 J/K·mol / 27.5 J/K·mol
= 1.302

Substituting the values of T1, T2, and γ into the equation, we can solve for V2.
200 K * (24.84 L)^(1.302-1) = 250 K * V2^(1.302-1)
(24.84 L)^(0.302) = (250 K / 200 K) * V2^(0.302)
(24.84 L)^(0.302) = (5/4) * V2^(0.302)
V2^(0.302) = ((24.84 L)^(0.302)) * (4/5)
V2 = ((24.84 L)^(0.302)) * (4/5)^(1/0.302)
V2 ≈ 31.84 L

3. Calculation of Work Done (W):
Since the process is reversible, we can use the equation for work done in an adiabatic process.
W = (P2 * V2 - P1 * V1) / (γ - 1)

Substituting the values of P1, P2, V1, V2, and γ into the equation, we can calculate the work done.
W = (2 atm * 31.84 L - 2 atm * 24.84 L) / (1.302 - 1)
W = (63