Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external pressure 4.3 MPa. The heat transferred in this process is _________ kJ mol⁻¹. (Rounded off to the nearest integer) [Use R = 8.314 J mol⁻¹K⁻¹]

Solution

Chemistry ma’am · 4 steps

Isothermal ideal gas means Δ U = 0, so the heat in is exactly the work done against Pₑₓₜ.

1. First law. For an ideal gas, internal energy depends only on temperature. At constant T, Δ U = 0, so q = -w. The expansion is against a constant external pressure, so w = -PₑₓₜΔ V and q = PₑₓₜΔ V.

   q = Pₑₓₜ nRT(1/P₂-1/P₁)

2. Volumes from pressures. Use V = nRT/P for both states. The pressure units cancel, so the bracket times Pₑₓₜ is 4.3(1/1.3-1/2.1) = 4.3 × 0.8/2.73 = 1.260, a pure number.
3. Put in numbers. nRT = 5 × 8.314 × 293 = 12180 J. So q = 1.260 × 12180 = 15348 J ≈ 15.35 kJ.
4. So. The heat transferred is about 15.35 kJ for the 5 moles, which rounds to 15.

The common slip: 3. Dividing by the 5 moles gives about 3.07, so 3. The answer key takes the total heat for all five moles, which is 15.

The idea to keep. Isothermal ideal gas: Δ U = 0, so q = -w. Against a fixed external pressure, write Δ V using V = nRT/P and let the pressure units cancel.

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