Exemplar Problems

Problem 15 : Calculate the work done (in joules) when 2 moles of an ideal gas expand isothermally from a volume of 5 liters to a volume of 10 liters at 300 K. (R = 8.314 J/(mol•K))

Solution :

The work done (W) during an isothermal expansion of an ideal gas is given by the formula: W = -nRT ln(Vf/Vi)

Where:

• n is the number of moles of the gas (2 moles).
• R is the ideal gas constant (8.314 J/(mol•K)).
• T is the temperature in Kelvin (300 K).
• Vi and Vf are the initial and final volumes, respectively (Vi = 5 liters, Vf = 10 liters).

W = -2 moles * 8.314 J/(mol•K) * 300 K * ln(10/5) W = -2 * 8.314 J/K * 300 K * ln(2) W ≈ -2 * 8.314 J/K * 300 K * 0.693 W ≈ -4,929.39 J ≈ -4.93 kJ

So, the work done during the isothermal expansion is approximately -4.93 kJ.