If three moles of an ideal gas at 300 K expand isothermally from 30 dm³ to 45 dm³ against a constant opposing pressure of 80 kPa, then the amount of heat transferred is _______ J.

Solution

Chemistry ma’am · 4 steps

Isothermal ideal gas means ΔU is zero, so the heat absorbed equals the work done against 80 kPa.

1. Internal energy. The internal energy of an ideal gas depends only on temperature. The temperature stays at 300 K, so Δ U = 0 and the first law Δ U = q + w gives q = -w.
2. Irreversible work. The gas pushes against a constant external pressure, so this is an irreversible expansion. The work is pₑₓₜΔ V, not nRT ln(V₂/V₁).

   q = -w = pₑₓₜ(V₂ - V₁)

3. Put in numbers. Δ V = 45 - 30 = 15 dm³ = 0.015 m³ and pₑₓₜ = 80 kPa = 8 × 10⁴ Pa. So q = 8 × 10⁴ × 0.015 = 1200 J.
4. So. The gas absorbs 1200 J of heat. All of it goes into the work of expansion.

The common slip: 3034. Using the reversible formula nRT ln(V₂/V₁) gives about 3034 J. The gas expands against a constant 80 kPa, so the work is pₑₓₜΔ V = 1200 J.

The idea to keep. For an ideal gas at constant temperature, Δ U = 0 and q = -w. Constant external pressure means pₑₓₜΔ V. Keep nRT ln(V₂/V₁) for reversible changes only.

Answer from the NTA final answer key. Solution by PrepWiser.