According to question , mole = 2.8 / 14 = 0.2 ; given value , T = 300k , P = 20 atm , N = 0.2 mole , now we use ideal gas equation , pv = nrt ;; 20 × v1 = 0.2 × 0.0821 × 300k ; v1 = 0.2463 L ( then we Calculate get value ) now , given external pressure 1atm , then we use ideal gas equation , pv = nrt ;; 1 × v2 = 0.2 × 0.0821 × 300 k ; v2 = 4.926 L ,,, now, we Calculate w = ? now , w = - p v ( because work is done against constant pressure ) ,, w = -1 × 0.2463 - 4.926 L ;; w = 4.6797 × 101.3 j ( 1 L atm = 101.33 ) ;; w= 474.05 j

Introduction
In this scenario, we are calculating the work done (W) during the isothermal expansion of nitrogen gas (N2) at 300 K and 20 atm against an external pressure of 1 atm.
Given Data:
- Mass of N2 = 2.8 g
- Molar mass of N2 = 28 g/mol
- Temperature (T) = 300 K
- Initial pressure (P1) = 20 atm
- External pressure (Pext) = 1 atm
Calculating Moles of N2:
- Moles of N2 = Mass / Molar mass
- Moles of N2 = 2.8 g / 28 g/mol = 0.1 mol
Work Calculation:
For isothermal expansion, the work done by the gas can be calculated using the formula:
W = -Pext * ΔV
Here, ΔV can be obtained from the ideal gas equation:
- P1V1 = nRT
- V1 = nRT / P1
Substituting the values:
- V1 = (0.1 mol) * (0.0821 L·atm/K·mol) * (300 K) / (20 atm)
- V1 = 1.23 L
Now, since the gas expands against an external pressure of 1 atm, we calculate the work done:
- W = -Pext * ΔV
- W = -1 atm * (Vfinal - Vinitial)
Since the gas can expand to an infinite volume, we consider W = -nRT ln(P1/P2), where P2 is the final pressure (1 atm):
- W = -(0.1 mol) * (0.0821 L·atm/K·mol) * (300 K) * ln(20/1)
- W ≈ - (2.46 L·atm) * 2.9957 = -7.36 L·atm
Conclusion:
The work done (W) during the isothermal expansion of N2 gas is approximately -7.36 L·atm, indicating that energy is being expended by the system.