2.2 g of nitrous oxide (N O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causi — Thermodynamics and Thermochemistry Chemistry Question
Question
2.2 g of nitrous oxide (N O) 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, ∆U is ‘–x’ J. The value of ‘x’ is _____. [Nearest integer] (Given: atomic mass of N = 14 g mol and of O = 16 g mol . Molar heat capacity of N O is 100 J K mol ) 2 –1 –1 2 –1 –1
💡 Solution & Explanation
**Step 1: Calculate moles of N₂O** Molar mass of N₂O = 2(14) + 16 = 44 g/mol Number of moles = 2.2 g ÷ 44 g/mol = 0.05 mol **Step 2: Calculate work done by the gas** Using the first law of thermodynamics: ΔU = q – W For work at constant pressure: W = P·ΔV ΔV = 167.75 mL – 217.1 mL = –49.35 mL = –49.35 × 10⁻⁶ m³ P = 1 atm = 101,325 Pa W = 101,325 Pa × (–49.35 × 10⁻⁶ m³) = –5.00 J **Step 3: Calculate heat absorbed** Using: q = n·Cp·ΔT q = 0.05 mol × 100 J/(mol·K) × (270 – 310) K q = 0.05 × 100 × (–40) = –200 J **Step 4: Calculate change in internal energy** ΔU = q – W ΔU = –200 – (–5.00) = –200 + 5.00 = –195 J **Step 5: Identify x** Since ΔU = –x J, we have: –195 = –x x = 195 Therefore, the answer is 195.00.