1. **State the problem:** We have an adiabatic process where pressure $P$ and volume $V$ satisfy the relation $$PV^{1.4} = C,$$ where $C$ is a constant. Given: - $V = 340$ cubic centimeters - $P = 85$ kPa - Pressure is decreasing at a rate of $\frac{dP}{dt} = -7$ kPa/min We need to find the rate of change of volume $\frac{dV}{dt}$ at this instant. 2. **Write the formula and differentiate:** The relation is $$P V^{1.4} = C.$$ Since $C$ is constant, differentiate both sides with respect to time $t$: $$\frac{d}{dt}(P V^{1.4}) = 0.$$ Using the product rule: $$\frac{dP}{dt} V^{1.4} + P \frac{d}{dt}(V^{1.4}) = 0.$$ 3. **Differentiate $V^{1.4}$:** $$\frac{d}{dt}(V^{1.4}) = 1.4 V^{0.4} \frac{dV}{dt}.$$ So the equation becomes: $$\frac{dP}{dt} V^{1.4} + P (1.4 V^{0.4} \frac{dV}{dt}) = 0.$$ 4. **Solve for $\frac{dV}{dt}$:** $$P (1.4 V^{0.4} \frac{dV}{dt}) = - \frac{dP}{dt} V^{1.4}$$ $$\frac{dV}{dt} = - \frac{\frac{dP}{dt} V^{1.4}}{1.4 P V^{0.4}} = - \frac{\frac{dP}{dt} V^{1.4}}{1.4 P V^{0.4}}.$$ Simplify powers of $V$: $$\frac{dV}{dt} = - \frac{\frac{dP}{dt} V^{\cancel{1.4}}}{1.4 P V^{\cancel{0.4}}} = - \frac{\frac{dP}{dt} V^{1.4 - 0.4}}{1.4 P} = - \frac{\frac{dP}{dt} V^{1.0}}{1.4 P} = - \frac{\frac{dP}{dt} V}{1.4 P}.$$ 5. **Plug in the values:** $$\frac{dV}{dt} = - \frac{-7 \times 340}{1.4 \times 85} = \frac{7 \times 340}{1.4 \times 85}.$$ Calculate denominator: $$1.4 \times 85 = 119.$$ Calculate numerator: $$7 \times 340 = 2380.$$ So: $$\frac{dV}{dt} = \frac{2380}{119} = 20.$$ 6. **Interpretation:** The volume is increasing at a rate of 20 cubic centimeters per minute at the given instant. **Final answer:** $$\boxed{20 \text{ cubic centimeters per minute}}.$$