Let's solve each part step by step. --- ### **Given Data** - **State A:** \( P_A = (1+\alpha) \times 10^5 \) Pa, \( V_A = 0.001 \) m³, \( \alpha = 6 \) So, \( P_A = 7 \times 10^5 \) Pa, \( V_A = 0.001 \) m³ - **State B:** \( V_B = 0.008 \) m³ - **Internal Energy:** \( U = DPV + \text{constant} \), where \( D = 1 + \beta \), \( \beta = 5 \) ⇒ \( D = 6 \) --- ## **a) Adiabatic Curve in the \( P \)-\( V \) Plane** For an adiabatic process: \( dU = -PdV \) (since \( Q = 0 \)). Given: \( U = DPV + \text{constant} \) Take the differential: \[ dU = D(PdV + VdP) \] For the adiabatic process: \[ dU = -PdV \] So, \[ D(PdV + VdP) = -PdV \] \[ DPdV + DVdP = -PdV \] \[ DPdV + DVdP + PdV = 0 \] \[ (DP + P)dV + DVdP = 0 \] \[ (D+1)P dV + D V dP = 0 \] \[ \frac{(D+1)}{D} \frac{dV}{V} + \frac{dP}{P} = 0 \] \[ \frac{dP}{P} = -\frac{D+1}{D} \frac{dV}{V} \] Integrate both sides: \[ \int \frac{dP}{P} = -\frac{D+1}{D} \int \frac{dV}{V} \] \[ \ln P = -\frac{D+1}{D} \ln V + C \] Exponentiate: \[ P V^{(D+1)/D} = \text{constant} \] For \( D = 6 \): \[ P V^{7/6} = \text{constant} \] **Final Adiabatic Curve:** \[ \boxed{P V^{7/6} = \text{constant}} \] --- ## **b) Final Pressure for Adiabatic Expansion A → B** Use \( P_A V_A^{7/6} = P_B V_B^{7/6} \): - \( P_A = 7 \times 10^5 \) Pa - \( V_A = 0.001 \) m³ - \( V_B = 0.008 \) m³ Solve for \( P_B \): \[ P_B = P_A \left(\frac{V_A}{V_B}\right)^{7/6} \] \[ P_B = 7 \times 10^5 \left(\frac{0.001}{0.008}\right)^{7/6} \] First, calculate the ratio: \[ \frac{0.001}{0.008} = 0.125 \] Then, raise to \( 7/6 \): \[ 0.125^{7/6} = (0.125)^{1.1667} \] \[ \ln(0.125) = \ln(1/8) = -2.0794 \] \[ -2.0794 \times 1.1667 = -2.427 \] \[ e^{-2.427} \approx 0.0884 \] So, \[ P_B = 7 \times 10^5 \times 0.0884 = 61,880 \text{ Pa} \approx \boxed{6.19 \times 10^4 \text{ Pa}} \] --- ## **c) \( Q \) and \( W \) for the Linear Process** **Linear Process:** \( P \) decreases linearly as \( V \) increases from \( A \) to \( B \). Let the equation be: \[ P = mV + c \] At \( V_A = 0.001 \), \( P_A = 7 \times 10^5 \) At \( V_B = 0.008 \), \( P_B \) (from the linear process, not the adiabatic one; but not given directly—assume it ends at some \( P_B \)). Let's denote \( P_B' \) for the linear process. However, with only one point, let's assume the pressure drops to zero (a common assumption if not specified), or you can leave the answer in terms of \( P_B' \). **Work Done (\( W \))**: \[ W = \int_{V_A}^{V_B} P\,dV \] For a linear \( P \) vs \( V \): \[ W = \frac{1}{2} (P_A + P_B') (V_B - V_A) \] But if \( P_B' \) is not specified, leave it as is. **Change in Internal Energy (\( \Delta U \)):** \[ U = DPV \] So, \[ \Delta U = D(P_B' V_B - P_A V_A) \] **First Law:** \( Q = \Delta U + W \) Plug values (with \( D = 6 \)): \[ W = \frac{1}{2}(P_A + P_B') (V_B - V_A) \] \[ \Delta U = 6(P_B' V_B - P_A V_A) \] \[ Q = \Delta U + W \] If \( P_B' = 0 \): \[ W = \frac{1}{2}(7 \times 10^5 + 0)(0.008 - 0.001) = \frac{1}{2}(7 \times 10^5)(0.007) = 2450 \text{ J} \] \[ \Delta U = 6[0 - 7 \times 10^5 \times 0.001] = -4200 \text{ J} \] \[ Q = -4200 + 2450 = -1750 \text{ J} \] If not, leave in terms of \( P_B' \). --- ## **d) Work Done in the Adiabatic Process** \[ W = \int_{V_A}^{V_B} P\,dV \] Recall: \( P V^{7/6} = \text{constant} \Rightarrow P = C V^{-7/6} \) So, \[ W = \int_{V_A}^{V_B} C V^{-7/6} dV \] Integrate: \[ \int V^{-7/6} dV = \frac{V^{-7/6 + 1}}{-7/6 + 1} = \frac{V^{-1/6}}{-(1/6)} = -6 V^{-1/6} \] So, \[ W = C \left[-6 V^{-1/6}\right]_{V_A}^{V_B} \] But \( C = P_A V_A^{7/6} \): \[ W = P_A V_A^{7/6} \left[ -6(V_B^{-1/6} - V_A^{-1/6}) \right] \] \[ W = -6 P_A V_A^{7/6} (V_B^{-1/6} - V_A^{-1/6}) \] Plug in the values: - \( P_A = 7 \times 10^5 \) - \( V_A = 0.001 \) - \( V_B = 0.008 \) Calculate \( V_A^{7/6} \): \[ V_A^{7/6} = (0.001)^{7/6} = 10^{-3 \times 7/6} = 10^{-3.5} = 3.16 \times 10^{-4} \] Now \( V_B^{-1/6} \): \[ V_B^{-1/6} = (0.008)^{-1/6} \] \[ \ln(0.008) = -4.8283 \] \[ -4.8283 \times (-1/6) = 0.8047 \] \[ e^{0.8047} = 2.236 \] \( V_A^{-1/6} \): \[ (0.001)^{-1/6} = 10^{-3 \times (-1/6)} = 10^{0.5} = 3.162 \] So, \[ W = -6 \times 7 \times 10^5 \times 3.16 \times 10^{-4} \times (2.236 - 3.162) \] \[ = -6 \times 7 \times 10^5 \times 3.16 \times 10^{-4} \times (-0.926) \] Calculate step by step: - \( 7 \times 10^5 \times 3.16 \times 10^{-4} = 7 \times 3.16 \times 10^{1} = 22.12 \times 10 = 221.2 \) - \( -6 \times 221.2 \times (-0.926) = -1327.2 \times (-0.926) = 1228.0 \) J **Final Work Done:** \[ \boxed{W \approx 1.23 \times 10^3\, \text{J}} \] --- ## **Summary Table** | Part | Final Answer | |------|-------------| | a | \( \boxed{P V^{7/6} = \text{constant}} \) | | b | \( \boxed{P_B \approx 6.19 \times 10^4\, \text{Pa}} \) | | c | \( \boxed{W = 2450\,\text{J},\ \Delta U = -4200\,\text{J},\ Q = -1750\,\text{J}} \) (if \(P_B' = 0\)) | | d | \( \boxed{W \approx 1.23 \times 10^3\, \text{J}} \) | Let me know if you need further clarifications or if the final pressure for the linear process (\(P_B'\)) is specified!