# Solution: Ideal Gas Problem ## Given Data - Number of moles, \( n = .8 \) mol - Initial Temperature, \( T_1 = 300 \) K - Final Temperature, \( T_2 = 450 \) K - Volume, \( V = 12 \) L \( = .012 \) m³ (since \(1 \text{ L} = .001 \text{ m}^3\)) - Universal Gas Constant, \( R = .0821 \) L·atm·K\(^{-1}\)·mol\(^{-1}\) (or \(8.314\) J·K\(^{-1}\)·mol\(^{-1}\)) --- ## Step 1: Calculate Initial Pressure (\( P_1 \)) Use the Ideal Gas Law: \[ P_1 V = nRT_1 \] \[ P_1 = \frac{nRT_1}{V} \] Substitute in values (using L·atm units): \[ P_1 = \frac{.8 \times .0821 \times 300}{12} \] \[ P_1 = \frac{.8 \times 24.63}{12} \] \[ P_1 = \frac{19.704}{12} \] \[ P_1 \approx 1.642 \text{ atm} \] --- ## Step 2: Calculate Final Pressure (\( P_2 \)) Since the volume is constant and only temperature changes, use the direct relationship from the ideal gas law: \[ \frac{P_2}{T_2} = \frac{P_1}{T_1} \] \[ P_2 = P_1 \times \frac{T_2}{T_1} \] \[ P_2 = 1.642 \times \frac{450}{300} \] \[ P_2 = 1.642 \times 1.5 \] \[ P_2 = 2.463 \text{ atm} \] --- ## Step 3: Work Done by the Gas For a rigid container (constant volume), the work done by the gas is: \[ W = P \Delta V = \] **No work is done, since the volume does not change.** --- ## Final Answers - **Initial Pressure**: \( \boxed{1.64 \text{ atm}} \) (rounded to two decimals) - **Final Pressure**: \( \boxed{2.46 \text{ atm}} \) (rounded to two decimals) - **Work Done by Gas**: \( \boxed{} \) (since volume is constant) --- ### Summary Table | Quantity | Value | |--------------------|---------------| | Initial Pressure | 1.64 atm | | Final Pressure | 2.46 atm | | Work Done (\( W \))| |