(1 pt) When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant.
Suppose that at a certain instant the volume is 390 cm3, and the pressure is 75 kPa (kPa = kiloPascals) and is decreasing at a rate of 8 kPa/minute. At what rate is the volume increasing at this instant?
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To find the rate at which the volume is increasing at a certain instant, we need to use the given information and apply the chain rule of differentiation. Let's break down the problem step by step:
Step 1: Given information:
- Initial volume V₁ = 390 cm³
- Initial pressure P₁ = 75 kPa
- Rate of change of pressure dP/dt = -8 kPa/minute (negative sign indicates decreasing pressure)
- Equation relating pressure and volume: PV¹.⁴ = C, where C is a constant.
Step 2: Applying the chain rule:
To find the rate at which the volume is increasing (dV/dt) at the given instant, we need to express dV/dt in terms of dP/dt. Since the equation relates both P and V, we can differentiate it with respect to time t using the chain rule.
Differentiating the equation PV¹.⁴ = C with respect to time t, we get:
V¹.⁴ dP/dt + P(1.4V⁰.⁴ dV/dt) = 0
Step 3: Solve for dV/dt:
Now we have an equation involving dV/dt that we can solve for. Rearranging the equation, we get:
1.4P(V⁻⁰.⁶)dV/dt = - V¹.⁴ dP/dt
dV/dt = - (V¹.⁴ dP/dt) / (1.4P(V⁻⁰.⁶))
Step 4: Substitute the given values:
Substitute the given values (V = 390 cm³, P = 75 kPa, dP/dt = -8 kPa/minute) into the equation we derived in step 3:
dV/dt = - (390¹.⁴ * -8) / (1.4 * 75 * 390⁻⁰.⁶)
Simplifying this expression will give us the rate at which the volume is increasing at the given instant.