Solution:

To find the work performed during the isothermal and reversible expansion of 2 moles of hydrogen gas, we can use the following equation:

Work (W) = -nRT ln(Vf/Vi)

Where:
- 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
- Vi is the initial volume
- Vf is the final volume

Given:
- n = 2 moles
- T = 25°C = 298 K
- Vi = 15 L
- Vf = 50 L

Calculating the Work:

Substituting the given values into the equation, we have:

W = -2 * 8.314 J/(mol·K) * 298 K * ln(50 L/15 L)

Simplifying the expression, we get:

W ≈ -2 * 8.314 J/(mol·K) * 298 K * ln(3.33)

W ≈ -2 * 8.314 J/(mol·K) * 298 K * 1.203

W ≈ -5980.6 J

Therefore, the work performed during the isothermal and reversible expansion of 2 moles of hydrogen gas from 15 L to 50 L is approximately -5980.6 J.

Explanation:

- The work performed during the expansion of a gas is given by the equation W = -nRT ln(Vf/Vi).
- In this case, we have 2 moles of hydrogen gas undergoing an isothermal and reversible expansion.
- The ideal gas constant, R, is a constant value.
- The temperature, T, is given as 25°C, which is converted to Kelvin by adding 273 (298 K).
- The initial volume, Vi, is given as 15 L.
- The final volume, Vf, is given as 50 L.
- Substituting these values into the equation, we can calculate the work performed.
- The natural logarithm (ln) of the ratio of final volume to initial volume determines the magnitude and direction of the work.
- The negative sign in the equation indicates that work is done on the system during expansion.
- The final answer is approximately -5980.6 J, indicating that work is done on the system.