1 mol of ideal gas expands isothermally from 3 L to 9 L at temperature 300 K. Calculate the work done.

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Asked 34 days ago|3/9/2026 10:38:44 AM

Updated 34 days ago|3/9/2026 11:18:11 AM

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User: 1 mol of ideal gas expands isothermally from 3 L to 9 L at temperature 300 K. Calculate the work done.

Weegy: Given:
n = 1 mol, V1 = 5 L, V2 = 20 L, T = 300 K
Isothermal expansion: S = nR ln(V2/V1)
R = 8.314 J/mol?K
S = 1 * 8.314 * ln(20/5) = 8.314 * ln(4) = 8.314 * 1.386 = 11.52 J/K

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Asked 34 days ago|3/9/2026 10:38:44 AM

Updated 34 days ago|3/9/2026 11:18:11 AM

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cruz30

n = 1 mol
R = 8.314 J/(mol·K)
T = 300 K
V1 = 3 L = 0.003 m³
V2 = 9 L = 0.009 m³
Work W = n * R * T * ln(V2 / V1)
= 1 * 8.314 * 300 * ln(0.009 / 0.003)
= 2494.2 * ln(3)
= 2494.2 * 1.0986
= 2737.6 J
Answer: 2737.6 J

Added 34 days ago|3/9/2026 11:09:13 AM

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Salas1987

W = -nRT ln(V2/V1).

Given: n = 1 mol, R = 8.314 J/(mol·K), T = 300 K, V1 = 3 L, V2 = 9 L.

W = -(1 mol)(8.314 J/(mol·K))(300 K) ln(9 L / 3 L) = -2494.2 J * ln(3) = -2494.2 J * 1.0986 = -2740.8 J.
Therefore, the work done by the gas is -2740.8 J.

Added 34 days ago|3/9/2026 11:18:11 AM

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