Problem 1: Ideal Gas Law — Volume Calculation

Question: Calculate the volume occupied by 2 mol of an ideal gas at 27°C and 2 atm.

Solution: Step 1 — Convert temperature to Kelvin: $T = 27 + 273 = 300 \text{ K}$

Step 2 — Identify known variables: $n = 2 \text{ mol}, \quad P = 2 \text{ atm}, \quad T = 300 \text{ K}, \quad R = 0.0821 \text{ L·atm/(mol·K)}$

Step 3 — Apply ideal gas equation: $V = \frac{nRT}{P} = \frac{2 \text{ mol} \times 0.0821 \frac{\text{L·atm}}{\text{mol·K}} \times 300 \text{ K}}{2 \text{ atm}}$

Step 4 — Unit check: $\frac{\text{mol} \times \frac{\text{L·atm}}{\text{mol·K}} \times \text{K}}{\text{atm}} = \frac{\text{L·atm}}{\text{atm}} = \text{L} \checkmark$

Step 5 — Calculate: $V = \frac{2 \times 0.0821 \times 300}{2} = \frac{49.26}{2} = \boxed{24.63 \text{ L}}$

Verification: At STP (0°C, 1 atm), 2 mol would occupy 2 × 22.4 = 44.8 L. At 300 K (higher T) and 2 atm (higher P), the effects partially cancel. Final 24.63 L < 44.8 L → higher P effect dominated. ✓