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
The formula for work done during a reversible isothermal expansion of an ideal gas is:
W = -nRT ln(V₂/V₁)
Where:

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

Step 2: Calculate the number of moles (n)
Using the ideal gas law, PV = nRT, with P₁ = 2 atm, V₁ = 2 L, T = 300 K, and 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
Using 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 ≈ -280.913 J

Step 5: Interpret the result
The negative sign indicates work done by the gas. The magnitude of the work done by the gas is approximately:
|W| ≈ 280.91 J
Rounded to three significant figures: 281 J

Step 6: Verification with alternative units
Using 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
Converting to joules (1 atm·L = 101.325 J):
W ≈ -2.773 × 101.325 ≈ -280.97 J
The magnitude is confirmed: ≈ 281 J

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