1. **State the problem:** We have a reversible no-flow process where the work done is $-156.2$ kJ, initial volume $V_1 = 0.845$ m³, and pressure varies with volume as $p = -730V + 690$ kPa. We need to find the final volume $V_2$. 2. **Formula for work in a reversible process:** Work done is the area under the $p$-$V$ curve, given by $$W = \int_{V_1}^{V_2} p \, dV$$ 3. **Substitute the pressure relation:** $$W = \int_{V_1}^{V_2} (-730V + 690) \, dV$$ 4. **Integrate:** $$W = \left[-730 \frac{V^2}{2} + 690V \right]_{V_1}^{V_2} = \left[-365V^2 + 690V \right]_{V_1}^{V_2}$$ 5. **Apply limits:** $$W = (-365V_2^2 + 690V_2) - (-365V_1^2 + 690V_1)$$ 6. **Plug in known values:** $$-156.2 = (-365V_2^2 + 690V_2) - (-365(0.845)^2 + 690(0.845))$$ Calculate the known term: $$-365(0.845)^2 + 690(0.845) = -365 \times 0.714025 + 583.05 = -260.62 + 583.05 = 322.43$$ 7. **Rewrite equation:** $$-156.2 = (-365V_2^2 + 690V_2) - 322.43$$ Add $322.43$ to both sides: $$-156.2 + 322.43 = -365V_2^2 + 690V_2$$ $$166.23 = -365V_2^2 + 690V_2$$ 8. **Rearrange to standard quadratic form:** $$-365V_2^2 + 690V_2 - 166.23 = 0$$ Multiply both sides by $\cancel{-1}$: $$\cancel{-}365V_2^2 + \cancel{-}690V_2 + \cancel{-}166.23 = 0$$ $$365V_2^2 - 690V_2 + 166.23 = 0$$ 9. **Solve quadratic equation:** Use quadratic formula: $$V_2 = \frac{690 \pm \sqrt{(-690)^2 - 4 \times 365 \times 166.23}}{2 \times 365}$$ Calculate discriminant: $$690^2 = 476100$$ $$4 \times 365 \times 166.23 = 242638.2$$ $$\sqrt{476100 - 242638.2} = \sqrt{233461.8} = 483.22$$ 10. **Calculate roots:** $$V_2 = \frac{690 \pm 483.22}{730}$$ Two possible values: $$V_2 = \frac{690 + 483.22}{730} = \frac{1173.22}{730} = 1.606$$ $$V_2 = \frac{690 - 483.22}{730} = \frac{206.78}{730} = 0.283$$ 11. **Interpretation:** Since $V_1 = 0.845$ m³ and work is negative (work done by the system), volume likely decreases, so **Final answer:** $$V_2 = 0.283 \text{ m}^3$$