Given:
- Initial temperature (T1) = 298 K
- Initial pressure (P1) = 15 bar
- Final pressure (P2) = 2.5 bar
- The process is adiabatic

To Find:
- Work done by the system during the process

Solution:

1. Calculating the final temperature (T2)
Since the process is adiabatic, the change in internal energy (ΔU) during the process is zero.

ΔU = 0

Using the first law of thermodynamics,

ΔU = Q - W

Where Q is the heat transferred and W is the work done.

Since Q = 0 (adiabatic process), ΔU = -W

ΔU = (3/2) nR (T2 - T1)

Where n is the number of moles of the gas and R is the gas constant.

Since the process is adiabatic, the number of moles and gas constant are constant. Therefore,

T2 = (2/3) T1 (P2/P1)^((γ-1)/γ)

Where γ is the ratio of specific heats (Cp/Cv) of the gas.

For an ideal gas, γ = Cp/Cv = 1.4

Substituting the given values, we get,

T2 = 158.39 K

2. Calculating the work done (W)
The work done by the system during an adiabatic process is given by,

W = (nCv)(T1 - T2)

Where Cv is the molar specific heat at constant volume.

For an ideal gas,

Cv = (3/2)R

Substituting the given values, we get

W = -3.22 kJ/mol

Therefore, the work done by the system during the process is -3.22 kJ/mol.

Explanation:
An adiabatic process is a thermodynamic process in which no heat is transferred to or from the system. During the process, the internal energy of the system remains constant. Therefore, the work done by the system is equal to the change in internal energy. In this problem, we used the first law of thermodynamics to calculate the final temperature of the gas. Using the final temperature and the initial temperature, we calculated the work done by the system during the process. The work done is negative, indicating that the system has done work on the surroundings.