Q: A piston contains 0.1 moles of an ideal gas at a pressure of 1 atm and a volume of 10 liters. The gas undergoes an isothermal expansion at a constant temperature of 300 K to a final volume of 20 liters. Calculate the work done by the gas during the expansion process and explain how it relates to the first law of thermodynamics.

Answer:

Given: Initial pressure, P₁ = 1 atm Initial volume, V₁ = 10 L Final volume, V₂ = 20 L Number of moles, n = 0.1 mol Temperature, T = 300 K

First, we can calculate the final pressure using the ideal gas law: P₁V₁ = nRT P₁ = nRT / V₁

P₂ = (0.1 mol * 0.0821 L.atm/mol.K * 300 K) / 20 L P₂ = 0.12375 atm

Now, we can use the equation for the work done during an isothermal process: W = -nRT * ln(V₂/V₁)

W = -(0.1 mol * 0.0821 L.atm/mol.K * 300 K) * ln(20/10) W ≈ -6.9 J

Explanation: The work done by the gas during the expansion process is calculated as -6.9 J. According to the first law of thermodynamics, the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W). In this case, since the expansion is isothermal (constant temperature), the change in internal energy is zero. Therefore, the work done by the gas is equal to the heat added to the system, and the negative sign indicates that the gas is doing work on the surroundings during the expansion process. This is in accordance with the first law of thermodynamics, which states that the total energy of an isolated system is constant, and energy can be transferred as either heat or work.