# Step-by-Step Solution ## Given Data - Number of moles, \( n = .8 \) mol - Initial volume, \( V = 12 \) L - Initial temperature, \( T_1 = 300 \) K - Final temperature, \( T_2 = 450 \) K - The container is rigid: \( \Delta V = \) (volume is constant) --- ## 1. Calculate Initial Pressure (\( P_1 \)) Use the ideal gas law: \[ PV = nRT \] Where \( R = .0821 \) L·atm/(mol·K) (using L and atm for convenience). \[ P_1 = \frac{nRT_1}{V} \] Plug in the values: \[ P_1 = \frac{.8 \times .0821 \times 300}{12} \] Calculate step-by-step: - \( .8 \times .0821 = .06568 \) - \( .06568 \times 300 = 19.704 \) - \( 19.704 / 12 = 1.642 \) So, \[ \boxed{P_1 = 1.64 \ \text{atm}} \] --- ## 2. Calculate Final Pressure (\( P_2 \)) Again, using the ideal gas law, but with \( T_2 = 450 \) K (same \( n \) and \( V \)): \[ P_2 = \frac{nRT_2}{V} \] \[ P_2 = \frac{.8 \times .0821 \times 450}{12} \] Calculate step-by-step: - \( .8 \times .0821 = .06568 \) - \( .06568 \times 450 = 29.556 \) - \( 29.556 / 12 = 2.463 \) So, \[ \boxed{P_2 = 2.46 \ \text{atm}} \] --- ## 3. Work Done by the Gas (\( W \)) Work done at constant volume (\( \Delta V = \)): \[ W = P \Delta V = \] Since the volume is constant, **no work is done**. \[ \boxed{W = } \] --- ## **Summary Table** | Quantity | Value | |-----------------------|---------------| | Initial Pressure | 1.64 atm | | Final Pressure | 2.46 atm | | Work Done by the Gas | | --- ## **Final Answers** - **Initial pressure (\( P_1 \)):** \( 1.64 \) atm - **Final pressure (\( P_2 \)):** \( 2.46 \) atm - **Work done (\( W \)):** \( \)

# Step-by-Step Solution ## Given Data - **Number of moles**: \( n = 0.8 \) mol - **Volume**: \( V = 12 \) L - **Initial temperature**: \( T_1 = 300 \) K - **Final temperature**: \( T_2 = 450 \) K - **Container characteristics**: Rigid (constant volume) --- ## 1. Calculate Initial Pressure (\( P_1 \)) Using the ideal gas law: \[ PV = nRT \] Rearranging gives: \[ P_1 = \frac{nRT_1}{V} \] Substituting known values: \[ P_1 = \frac{0.8 \times 0.0821 \times 300}{12} \] ### Calculation Steps 1. Calculate \( 0.8 \times 0.0821 \): \[ 0.8 \times 0.0821 = 0.06568 \] 2. Multiply by 300: \[ 0.06568 \times 300 = 19.704 \] 3. Divide by 12: \[ \frac{19.704}{12} = 1.642 \] Thus, \[ \boxed{P_1 = 1.64 \ \text{atm}} \] --- ## 2. Calculate Final Pressure (\( P_2 \)) Using the same ideal gas law for the final state: \[ P_2 = \frac{nRT_2}{V} \] Substituting known values: \[ P_2 = \frac{0.8 \times 0.0821 \times 450}{12} \] ### Calculation Steps 1. Calculate \( 0.8 \times 0.0821 \): \[ 0.8 \times 0.0821 = 0.06568 \] 2. Multiply by 450: \[ 0.06568 \times 450 = 29.556 \] 3. Divide by 12: \[ \frac{29.556}{12} = 2.463 \] Thus, \[ \boxed{P_2 = 2.46 \ \text{atm}} \] --- ## 3. Work Done by the Gas (\( W \)) At constant volume, the work done by the gas is calculated as: \[ W = P \Delta V \] Since the volume is constant (\( \Delta V = 0 \)), the work done is: \[ W = 0 \] Thus, \[ \boxed{W = 0} \] --- ## Summary Table | Quantity | Value | |-----------------------|----------------| | Initial Pressure | 1.64 atm | | Final Pressure | 2.46 atm | | Work Done by the Gas | 0 | --- ## Final Answers - **Initial pressure (\( P_1 \))**: \( 1.64 \) atm - **Final pressure (\( P_2 \))**: \( 2.46 \) atm - **Work done (\( W \))**: \( 0 \)