Let's solve each part step-by-step: #### **Given Data** - Initial temperature, \( T_0 = 20^\circ C = 293.15 \) K - Initial pressure, \( P_0 = 1 \) bar = \( 1 \times 10^5 \) Pa - Initial volume, \( V_0 = 8 \) cm³ = \( 8 \times 10^{-6} \) m³ - Expansivity, \( \beta = 5.0 \times 10^{-5} \) K⁻¹ - Bulk modulus, \( K = 1.5 \times 10^{11} \) Nm⁻² --- ### **a. Find change in volume if pressure is kept constant and temperature increases to 50°C** At constant pressure, the change in volume due to temperature is: \[ \Delta V = V_0 \beta \Delta T \] where - \(\Delta T = T_1 - T_0 = 50^\circ C - 20^\circ C = 30\) K Plug in the values: \[ \Delta V = 8 \times 10^{-6} \times 5.0 \times 10^{-5} \times 30 \] \[ \Delta V = 8 \times 10^{-6} \times 1.5 \times 10^{-3} \] \[ \Delta V = 1.2 \times 10^{-8} \text{ m}^3 \] Convert to cm³ (\(1 \text{ m}^3 = 10^6 \text{ cm}^3\)): \[ \Delta V = 1.2 \times 10^{-8} \times 10^6 = 1.2 \times 10^{-2} \text{ cm}^3 = 0.012 \text{ cm}^3 \] **Final Answer (a):** \[ \boxed{\Delta V = 0.012 \text{ cm}^3} \] --- ### **b. Find change in volume if \(T\) is constant and pressure is increased to \(P_1 = 5\) bar** At constant temperature, the change in volume due to pressure is: \[ \Delta V = -\frac{V_0}{K} \Delta P \] where \[ \Delta P = P_1 - P_0 = (5 - 1) \text{ bar} = 4 \text{ bar} = 4 \times 10^5 \text{ Pa} \] Plug in the values: \[ \Delta V = -\frac{8 \times 10^{-6}}{1.5 \times 10^{11}} \times 4 \times 10^5 \] \[ = -8 \times 10^{-6} \times \frac{4 \times 10^5}{1.5 \times 10^{11}} \] \[ = -8 \times 10^{-6} \times \frac{4}{1.5} \times 10^{-6} \] \[ = -8 \times 10^{-6} \times 2.6667 \times 10^{-6} \] \[ = -21.3336 \times 10^{-12} \] \[ = -2.13 \times 10^{-11} \text{ m}^3 \] Convert to cm³ (\(1 \text{ m}^3 = 10^6 \text{ cm}^3\)): \[ \Delta V = -2.13 \times 10^{-11} \times 10^6 = -2.13 \times 10^{-5} \text{ cm}^3 \] **Final Answer (b):** \[ \boxed{\Delta V = -2.1 \times 10^{-5} \text{ cm}^3} \] --- ### **c. Find change in pressure if \(V\) is kept constant and \(T_1 = 50^\circ C\)** At constant volume: \[ \Delta V = 0 = V_0 \beta \Delta T - \frac{V_0}{K} \Delta P \] Rearrange for \(\Delta P\): \[ 0 = V_0 \beta \Delta T - \frac{V_0}{K} \Delta P \] \[ \frac{V_0}{K} \Delta P = V_0 \beta \Delta T \] \[ \Delta P = K \beta \Delta T \] Plug in the values: \[ \Delta P = 1.5 \times 10^{11} \times 5.0 \times 10^{-5} \times 30 \] \[ = 1.5 \times 5.0 \times 30 \times 10^{11-5} \] \[ = 7.5 \times 30 \times 10^6 \] \[ = 225 \times 10^6 \text{ Pa} \] \[ = 2.25 \times 10^8 \text{ Pa} \] Convert to bar (\(1 \text{ bar} = 10^5 \text{ Pa}\)): \[ \Delta P = \frac{2.25 \times 10^8}{10^5} = 2250 \text{ bar} \] **Final Answer (c):** \[ \boxed{\Delta P = 2.25 \times 10^8 \text{ Pa} = 2250 \text{ bar}} \] --- Let me know if you need more explanation or calculations!

**Final Answers:** a. \(\boxed{0.012 \text{ cm}^3}\) (volume increase when temperature rises to 50°C at constant pressure) b. \(\boxed{-2.1 \times 10^{-5} \text{ cm}^3}\) (volume decrease when pressure increases to 5 bar at constant temperature) c. \(\boxed{2.25 \times 10^8 \text{ Pa} \text{ (or 2250 bar)}}\) (pressure increase when temperature rises to 50°C at constant volume)