2.2 g of nitrous oxide (N₂O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, ΔU is '-x' J. The value of 'x' is ________. [nearest integer]
(Given : atomic mass of N = 14 g mol⁻¹ and of O = 16 g mol⁻¹. Molar heat capacity of N₂O is 100 J K⁻¹ mol⁻¹)

Solution

Chemistry ma’am · 4 steps

Heat from nCΔ T, work from -PΔ V, then add them. Compression means w is positive.

1. Moles of gas. Molar mass of N₂O is 2 × 14 + 16 = 44 g mol⁻¹. So n = 2.2/44 = 0.05 mol.
2. Heat exchanged. The pressure is constant, so take the given heat capacity as Cₚ. Then q = nCₚΔ T = 0.05 × 100 × (270 - 310) = -200 J. Heat leaves the gas as it cools.
3. Work done. Δ V = 167.75 - 217.1 = -49.35 mL = -0.04935 L. So w = -PΔ V = +0.04935 L atm. With 1 L atm = 101.3 J, this is w ≈ +5 J, done on the gas.

   Δ U = q + w = -200 + 5 = -195 J

4. So. Δ U = -195 J, so x = 195.

The common slip: 205. Treating the compression work as negative gives -200 - 5 = -205, so 205. Work done on a compressed gas is positive.

The idea to keep. Use Δ U = q + w with w = -PΔ V. When a gas is compressed, work is done on it, so w is positive. Convert L atm to joules before you add.

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