Calculate the maximum work done when two moles of N2 expands isothermally from 10 litres to 20 litres at 25o C

W = nRT*ln(Vfinal/Vinitial)

If you want W in joules use R = 8.314

To calculate the maximum work done during an isothermal expansion, we need to apply the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) supplied to the system minus the work (W) done by the system:

ΔU = Q - W

In an isothermal process, the temperature remains constant. Therefore, the change in internal energy (ΔU) is equal to zero. This means that the heat supplied to the system is equal to the work done by the system:

Q = W

To calculate the work done, we can use the following equation for an ideal gas:

W = -nRT ln(V2/V1)

Where:
W is the work done
n is the number of moles of the gas
R is the ideal gas constant (8.314 J/mol·K)
T is the temperature in Kelvin
V1 and V2 are the initial and final volumes, respectively.

Given the information in the question, we have:
n = 2 moles
R = 8.314 J/mol·K
T = 25 degrees Celsius = 298.15 Kelvin
V1 = 10 liters
V2 = 20 liters

Substituting these values into the equation, we can calculate the work done:

W = -2 * 8.314 J/mol·K * 298.15 K * ln(20/10)

Now, let's solve this equation step by step:

W = -2 * 8.314 J/mol·K * 298.15 K * ln(2)

Using a natural logarithm calculator, we can find that ln(2) is approximately 0.6931.

W = -2 * 8.314 J/mol·K * 298.15 K * 0.6931

W ≈ -4141.267 J

Since work is a negative value in this case, it means that the system is doing work on the surroundings. The magnitude of the work in this case is approximately 4141.267 J.