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 530 cubic centimeters and the pressure is 97 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?
(Pa stands for Pascal -- it is equivalent to one Newton/(meter squared); kPa is a kiloPascal or 1000 Pascals. )
V^1.4 = C/P
1.4 V^.4 dV/dt = (C/P^2)dP/dt
1.4 (530)^.4 dV/dt = (C/97^2)(10)
get C from initial condition
C = 97 * 530^1.4
the answer is shown to be incorrect....
Actually he forgot to convert the kpa(1000 * N/m^2) into 1000 * N/cm^2
To solve this problem, we can start by differentiating the given equation with respect to time.
Differentiating both sides of the equation PV^1.4 = C with respect to time gives:
d/dt(PV^1.4) = d/dt(C)
Using the product rule, the left side becomes:
dP/dt * V^1.4 + P * dV/dt * 1.4V^0.4 = 0
Since we are given that the air is expanding adiabatically (without gaining or losing heat), we know that dP/dt is given as -10 kPa/minute (decreasing at a rate of 10 kPa/minute).
Substituting the given information into the equation, we have:
-10 * V^1.4 + P * dV/dt * 1.4V^0.4 = 0
Now we can substitute the given values at the certain instant:
V = 530 cubic centimeters
P = 97 kPa
dP/dt = -10 kPa/minute
Plugging in these values into the equation, we can solve for dV/dt, the rate at which the volume is increasing at this instant.
-10 * (530)^1.4 + 97 * dV/dt * 1.4 * (530)^0.4 = 0
Simplifying the equation gives:
-10 * 530^1.4 + 97 * dV/dt * 1.4 * 530^0.4 = 0
Solving for dV/dt gives:
dV/dt = (10 * 530^1.4) / (97 * 1.4 * 530^0.4)
Simplifying further:
dV/dt = 7.8979 cubic centimeters/minute
Therefore, the rate at which the volume is increasing at this instant is approximately 7.8979 cubic centimeters per minute.