When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV^1.4=C
where C is a constant. Suppose that at a certain instant the volume is 610 cubic centimeters and the pressure is 89 kPa and is decreasing at a rate of 10 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?

48.96 cm³/min

Step-by-step explanation:

We are given;

Relationship between pressure P and volume V; PV^(1.4) = C

Volume;V = 610 m³

Pressure; P = 89 KPa

Rate of decreasing pressure; dP/dt = -10 kPa/minute.

We want to find the rate at which the volume is increasing at that instance, thus, its means we need to find dV/dt

So, we will differentiate the relationship equation of P and V given.

Thus, we have;

[V^(1.4)(dP/dt)] + d(V^(1.4))/dt = dC/dt

Differentiating this gives us;

[(dP/dt) × (V^(1.4))] + [1.4 × P × V^(0.4) × (dV/dt)] = 0

Plugging in the relevant values, we have;

(-10 × 610^(1.4)) + (1.4 × 89 × 610^(0.4) × (dV/dt)) = 0

This gives;

-79334.44155 + 1620.5035(dV/dt) = 0

Rearranging, we have;

1620.5035(dV/dt) = 79334.44155

Divide both sides by 1620.5035 to give;

dV/dt = 79334.44155/1620.5035

dV/dt = 48.96 cm³/min

∠cbe and ∠deb are same-side interior angles.

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