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 340 cubic centimeters and the pressure is 87 kPa and is decreasing at a rate of 14 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
take the differential..
PV^1.4=0
1.4 P v^.4 dv/dt + V^1.4 dp/dt=0
Solve for dv/dt.
oopss. STart off with PV^1.4=C
typo.
what do i do with the 14kpa/min?
Plug in dp/dt =-14 kpa/min because it is decreasing at a rate. So solving for dv/dt= (1-(V^(1/4)dP/dt))/((1/4)PV^(-3/4)). Your answer so be
dv/dt= (1-((340)^(1/4)(-14))/((1/4)(87)(340)^(-3/4)).
To solve this problem, we can apply the chain rule from calculus.
Given the equation PV^1.4 = C, we need to differentiate both sides of this equation with respect to time.
Differentiating the left side of the equation:
d(PV^1.4)/dt = d(C)/dt
Using the product rule, we get:
V^1.4 * dP/dt + 1.4P * dV/dt = 0
We are given dP/dt = -14 kPa/minute (pressure is decreasing at a rate of 14 kPa/minute).
Also, we know P = 87 kPa and V = 340 cubic centimeters.
Plugging these values into the equation, we can solve for dV/dt, which represents the rate at which the volume is changing.
(340^1.4) * (-14) + 1.4 * 87 * dV/dt = 0
Simplifying this equation:
-14 * (340^1.4) + 1.4 * 87 * dV/dt = 0
Now, solve for dV/dt:
dV/dt = (14 * 340^1.4) / (1.4 * 87)
Calculating this expression, we find:
dV/dt ≈ 309.9 cubic centimeters per minute
Therefore, the volume is increasing at a rate of approximately 309.9 cubic centimeters per minute.