Question

The work done by a gas during an isothermal expansion from 2 L to 4 L at a constant temperature, with pressure initially at 2 atm, is: (Use R = 8.314 J/mol·K, T = 300 K).

• A. 281 J
• B. 831.4 J
• C. 1247.1 J
• D. 1662.8 J

Hint:

For isothermal processes, use the work formula with natural logarithm of volume ratio.

Answer 1

The Correct Option is A

Step 1: Identify the formula
For a reversible isothermal expansion of an ideal gas, the work done is given by:
W = -nRT ln(V₂/V₁)
where:

• n = number of moles
• R = gas constant (8.314 J/mol·K for joules, or 0.08206 L·atm/mol·K for atm·L)
• T = temperature (300 K)
• V₁ = initial volume (2 L)
• V₂ = final volume (4 L)

Step 2: Calculate the number of moles (n)
Use the ideal gas law: PV = nRT
Given: P₁ = 2 atm, V₁ = 2 L, T = 300 K, R = 0.08206 L·atm/mol·K
n = (P₁V₁)/(RT)
n = (2 atm × 2 L)/(0.08206 L·atm/mol·K × 300 K)
n = 4 / (24.618) ≈ 0.1625 mol

Step 3: Calculate the volume ratio
V₂/V₁ = 4 L / 2 L = 2
ln(V₂/V₁) = ln(2) ≈ 0.6931

Step 4: Compute the work done
Use R = 8.314 J/mol·K for work in joules:
W = -(0.1625 mol) × (8.314 J/mol·K) × (300 K) × ln(2)
W = -(0.1625) × (8.314) × (300) × (0.6931)
W ≈ -(0.1625) × 2494.2 × 0.6931 ≈ -280.913 J

Step 5: Interpret the result
The negative sign indicates work done by the gas (thermodynamic convention). For work done by the gas, we take the positive magnitude:
|W| ≈ 280.91 J
Rounded to three significant figures: 281 J

Step 6: Verification with alternative units
In atm·L units:
nRT = (0.1625 mol) × (0.08206 L·atm/mol·K) × (300 K) ≈ 4.001 atm·L
W = -4.001 × ln(2) ≈ -4.001 × 0.6931 ≈ -2.773 atm·L
Convert to joules (1 atm·L = 101.325 J):
W ≈ -2.773 × 101.325 ≈ -280.97 J
Magnitude confirms: ≈ 281 J

Final Answer: The work done by the gas is approximately 281 J.