Air expands adiabatically in accordance with the law PV^1.4=Const. If at a given time the volume is 14 cubic feet and the pressure is 40 pounds per square inch, at what rate is the pressure changing when the volume is decreasing 1 cubic foot per second?
To find the rate at which the pressure is changing when the volume is decreasing, we first need to calculate the corresponding initial value of the pressure.
Given:
Initial volume (V1) = 14 cubic feet
Initial pressure (P1) = 40 pounds per square inch
Power in the adiabatic expansion (n) = 1.4
The equation for the adiabatic expansion is given by the equation: PV^n = const.
Using this equation, we can calculate the initial constant:
P1V1^n = P2V2^n
Where:
P2 = final pressure (the pressure when the volume decreases)
V2 = final volume (the volume after decreasing 1 cubic foot)
n = power in the adiabatic expansion
Now let's calculate the final volume (V2):
V2 = V1 - ΔV
Where:
ΔV = change in volume = -1 cubic foot (volume decreases by 1 cubic foot per second)
Substituting the known values into the equation:
V2 = 14 - (-1) = 14 + 1 = 15 cubic feet
Now we can calculate the initial constant using the initial and final values of volume and pressure:
P1V1^n = P2V2^n
P1 * 14 * 1.4 = P2 * 15 * 1.4
40 * 14 * 1.4 = P2 * 15 * 1.4
Simplifying the equation gives:
P2 = (40 * 14 * 1.4) / (15 * 1.4)
P2 = 560 / 15
P2 ≈ 37.33 pounds per square inch (rounded to two decimal places)
Now, we can differentiate the equation PV^n = const with respect to time:
(dP/dt)V + (dV/dt)P^n = 0
Since we know that (dV/dt) = -1 cubic foot per second (volume decreasing) and P = P2, we can solve for (dP/dt):
(dP/dt)V + (-1)(37.33)^1.4 = 0
(dP/dt)V ≈ (37.33)^1.4
(dP/dt) ≈ [(37.33)^1.4]/V
Substituting the known values:
(dP/dt) ≈ [(37.33)^1.4]/14
(dP/dt) ≈ 3.2204 pounds per square inch per second (rounded to four decimal places)
Therefore, the rate at which the pressure is changing when the volume is decreasing 1 cubic foot per second is approximately 3.2204 pounds per square inch per second.
To find the rate at which the pressure is changing when the volume is decreasing, we need to use calculus and differentiate the given equation PV^1.4 = Const.
Let's start by differentiating the equation with respect to time t, using the chain rule:
d/dt (PV^1.4) = d/dt (Const)
To differentiate PV^1.4, we need to differentiate both terms separately:
dP/dt * V^1.4 + P * d(V^1.4)/dt = 0
Now, let's simplify the equation using the power rule of differentiation:
dP/dt * V^1.4 + 1.4P * V^0.4 * dV/dt = 0
Since we are interested in finding the rate of change of pressure (dP/dt) when the volume (V) is changing, we can rearrange the equation to solve for dP/dt:
dP/dt = - (1.4P * V^0.4 * dV/dt) / V^1.4
Now, we can substitute the given values into the equation to find the rate at which the pressure is changing:
P = 40 pounds per square inch (given)
V = 14 cubic feet (given)
dV/dt = -1 cubic foot per second (given)
Plugging these values into the equation, we have:
dP/dt = - (1.4 * 40 * 14^0.4 * (-1)) / 14^1.4
Calculating the expression, we find:
dP/dt ≈ 37.66 pounds per square inch per second
Therefore, the rate at which the pressure is changing when the volume is decreasing 1 cubic foot per second is approximately 37.66 pounds per square inch per second.
k = P V^1.4
k = 40 * 14^1.4
k = 1609
so
P = 1609 V^-1.4
dP/dt = 1609 (-1.4)(V^-2.4) dV/dt
= -2253 (14)^-2.4 * -1
= 4 psi/s