# Step-by-Step Solution We are given: - Initial pressure: \( p_1 = 20 \) lbf/in\(^2\) - Initial temperature: \( T_1 = 500 \) R - Initial volume: \( V_1 = 9 \) ft\(^3\) - Final volume: \( V2 = 1 \) ft\(^3\) - Process: \( pv^{1.1} = \text{constant} \) - Ideal gas behavior, use constant specific heats at \( T_1 \) - Find: **Work** \( W \) and **Heat Transfer** \( Q \) in Btu --- ## 1. **Convert Units Where Necessary** - \( 1 \) lbf/in\(^2\) = \( 144 \) lbf/ft\(^2\) - \( p_1 = 20 \times 144 = 288 \) lbf/ft\(^2\) --- ## 2. **Find Mass of Air** The ideal gas law: \[ p_1 V_1 = m R T_1 \] Where: - \( R = 53.35 \) ft·lbf/(lbm·R) for air \[ m = \frac{p_1 V_1}{R T_1} = \frac{288 \times 9}{53.35 \times 500} = \frac{25920}{26675} \approx .972 \ \text{lbm} \] --- ## 3. **Find Final Pressure \( p_2 \) Using Process Equation** Given \( p V^{1.1} = \text{constant} \): \[ \frac{p_2}{p_1} = \left(\frac{V_1}{V_2}\right)^{1.1} \] \[ p_2 = p_1 \left(\frac{V_1}{V_2}\right)^{1.1} = 288 \times \left(\frac{9}{1}\right)^{1.1} \] Calculate \( 9^{1.1} \): - \( \ln 9 = 2.197 \) - \( 1.1 \ln 9 = 2.417 \) - \( e^{2.417} \approx 11.22 \) \[ p_2 = 288 \times 11.22 \approx 32{,}314 \ \text{lbf/ft}^2 \] --- ## 4. **Find Final Temperature \( T_2 \)** Use ideal gas law: \[ \frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2} \implies T_2 = T_1 \frac{p_2 V_2}{p_1 V_1} \] \[ T_2 = 500 \times \frac{32,314 \times 1}{2,880 \times 9} = 500 \times \frac{32,314}{25,920} = 500 \times 1.247 = 623.5 \ \text{R} \] --- ## 5. **Work Done During the Process** For a polytropic process (\( pV^n = \text{const} \)), work is: \[ W = \frac{p_2 V_2 - p_1 V_1}{1 - n} \] Where \( n = 1.1 \). Plug in values: - \( p_1 V_1 = 288 \times 9 = 25,920 \) ft·lbf - \( p_2 V_2 = 32,314 \times 1 = 32,314 \) ft·lbf \[ W = \frac{32,314 - 25,920}{1 - 1.1} = \frac{6,394}{-.1} = -63,940 \ \text{ft·lbf} \] Convert ft·lbf to Btu: - \( 1 \) Btu = \( 778.17 \) ft·lbf \[ W = \frac{-63,940}{778.17} \approx -82.2 \ \text{Btu} \] --- ## 6. **Change in Internal Energy \( \Delta U \)** \[ \Delta U = m c_v (T_2 - T_1) \] For air, at \( T_1 = 500 \) R (\( 40^\circ \)F), \( c_v \approx .171 \) Btu/(lbm·R) \[ \Delta U = .972 \times .171 \times (623.5 - 500) = .972 \times .171 \times 123.5 \] \[ = .972 \times 21.125 = 20.54 \ \text{Btu} \] --- ## 7. **Heat Transfer \( Q \) (First Law of Thermodynamics)** \[ Q = \Delta U + W \] \[ Q = 20.54 + (-82.2) = -61.7 \ \text{Btu} \] --- ## 8. **Final Answers** **Work done by air (on surroundings):** \[ \boxed{W = -82.2 \ \text{Btu}} \] (Negative: work is done *on* the air) **Heat transfer to the air:** \[ \boxed{Q = -61.7 \ \text{Btu}} \] (Negative: heat is *leaving* the air) --- ## **Summary Table** | Quantity | Value | Units | |---------------|--------------|-------------| | Mass \( m \) | .972 | lbm | | Final Pressure \( p_2 \) | 32,314 | lbf/ft\(^2\) | | Final Temp \( T_2 \) | 623.5 | R | | Work \( W \) | -82.2 | Btu | | Heat \( Q \) | -61.7 | Btu |