500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are: Given: R = 8.3 J K⁻¹ mol⁻¹

Solution

Chemistry ma’am · 4 steps

The gas stays at 1 atm, so the heat is qₚ. Part of it does expansion work and the rest becomes Δ U.

1. Constant pressure. The argon is held at 1.00 atm, so it absorbs heat at constant pressure and expands. Argon is monatomic, so Cₚ = 5/2R = 20.75 J K⁻¹ mol⁻¹ and Cᵥ = 3/2R = 12.45 J K⁻¹ mol⁻¹.

   qₚ = nCₚΔ T ⇒ Δ T = 500/(0.5 × 20.75) ≈ 48.2 K

2. Final temperature. T = 298 + 48.2 ≈ 346 K. Among the options, the closest value is 348 K.
3. Internal energy. Δ U = nCᵥΔ T = Cᵥ/Cₚ × qₚ = 3/5 × 500 = 300 J. The other 200 J leaves as expansion work, nRΔ T = 0.5 × 8.3 × 48.2 ≈ 200 J.
4. So. Δ U = 300 J exactly, and the temperature comes out near 346 K. Only (b) pairs 300 J with a temperature close to that, so the answer is (b).

The idea to keep. First decide the path. At constant pressure, use q = nCₚΔ T to find Δ T, then Δ U = nCᵥΔ T. For a monatomic gas, Δ U is 3/5 of the heat supplied.

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