Let's solve each part step by step: ### Given: - **Pressure (P):** 1800 Pa (constant) - **Initial temperature (T₁):** 250 K - **Final temperature (T₂):** 750 K - **Initial volume (V₁):** 0.15 m³ (for parts a–c) - **Ideal Gas Law:** \( PV = nRT \) - **R (universal gas constant):** 8.314 J/(mol·K) --- ### **a. Final volume (V₂) when V₁ = 0.15 m³** Using the Ideal Gas Law, for a constant pressure process: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \implies V_2 = V_1 \frac{T_2}{T_1} \] Substitute values: \[ V_2 = 0.15\, \text{m}^3 \times \frac{750\, \text{K}}{250\, \text{K}} = 0.15\, \text{m}^3 \times 3 = 0.45\, \text{m}^3 \] **Final Volume, \( V_2 = 0.45 \) m³** --- ### **b. Change in volume (\( \Delta V \))** \[ \Delta V = V_2 - V_1 = 0.45\, \text{m}^3 - 0.15\, \text{m}^3 = 0.30\, \text{m}^3 \] **Change in Volume, \( \Delta V = 0.30 \) m³** --- ### **c. Work done by the gas (\( W \)) during expansion** For constant pressure: \[ W = P \Delta V \] \[ W = 1800\, \text{Pa} \times 0.30\, \text{m}^3 = 540\, \text{J} \] **Work Done, \( W = 540 \) J** --- ### **d. If initial volume is \(0.24\, \text{m}^3\)** Repeat steps a–c: #### **Final volume (\( V_2' \)):** \[ V_2' = V_1' \frac{T_2}{T_1} = 0.24\, \text{m}^3 \times \frac{750}{250} = 0.24\, \text{m}^3 \times 3 = 0.72\, \text{m}^3 \] #### **Change in volume (\( \Delta V' \)):** \[ \Delta V' = V_2' - V_1' = 0.72 - 0.24 = 0.48\, \text{m}^3 \] #### **Work done (\( W' \)):** \[ W' = P \Delta V' = 1800\, \text{Pa} \times 0.48\, \text{m}^3 = 864\, \text{J} \] **Work Done, \( W' = 864 \) J** --- ### **e. Is the same amount of gas involved in both situations?** Let’s check the number of moles using the ideal gas law at the initial state for both cases: \[ n = \frac{PV}{RT} \] #### **Case 1:** \[ n_1 = \frac{1800 \times 0.15}{8.314 \times 250} = \frac{270}{2078.5} \approx 0.13\, \text{mol} \] #### **Case 2:** \[ n_2 = \frac{1800 \times 0.24}{8.314 \times 250} = \frac{432}{2078.5} \approx 0.208\, \text{mol} \] **No, the amount of gas is not the same in both cases.** - In the first case, \( n_1 \approx 0.13 \) mol - In the second case, \( n_2 \approx 0.208 \) mol **Explanation:** The difference in initial volume (at the same P and T) means different amounts of gas are present in each scenario. --- ## **Summary Table** | Part | Value | Unit | Explanation | |------|-------|------|-------------| | a | 0.45 | m³ | Final volume (V₁ = 0.15 m³) | | b | 0.30 | m³ | Change in volume (ΔV) | | c | 540 | J | Work done by gas | | d | 0.72, 0.48, 864 | m³, m³, J | Final volume, ΔV, Work (V₁ = 0.24 m³) | | e | No | | Amount of gas not the same | If you need any deeper explanation for any step, let me know!