Problem Statement:
Calculate the amount of work done by 2 moles of an ideal gas at 298 K in a reversible isothermal expansion from 10 liters to 20 liters.

Solution:
To calculate the work done by an ideal gas during an isothermal expansion, we can use the formula:

W = -nRT ln(Vf/Vi)

Where:
W is the work done (in joules)
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 = 298 K
Vi = 10 liters
Vf = 20 liters

Calculation:
Substituting the given values in the formula, we get:

W = -2 * 8.314 J/(mol·K) * 298 K * ln(20/10)

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

W ≈ -2 * 8.314 J/(mol·K) * 298 K * 0.693

W ≈ -3898.4 J

Therefore, the amount of work done by the gas during the reversible isothermal expansion is approximately -3898.4 J. The negative sign indicates that work is done on the system, which is expected for an expansion.

Explanation:
- The work done by an ideal gas during an isothermal expansion can be calculated using the formula W = -nRT ln(Vf/Vi).
- In this formula, n represents the number of moles of the gas, R is the ideal gas constant, T is the temperature, Vi is the initial volume, and Vf is the final volume.
- Substituting the given values into the formula, we can calculate the work done.
- The natural logarithm ln(Vf/Vi) represents the ratio of the final volume to the initial volume.
- The negative sign in the formula indicates that work is done on the system during expansion.
- In this specific problem, we are given that the gas has 2 moles, the temperature is 298 K, the initial volume is 10 liters, and the final volume is 20 liters.
- Substituting these values into the formula, we can calculate the work done by the gas.
- The calculated value of work is approximately -3898.4 J, indicating that work is done on the system during the reversible isothermal expansion.

-3.4 KJ