Boyle’s law for the expansion of a gas is PV = k, where P is the pressure (in pounds per unit of area), V is the volume of the gas, and k is a constant.
Find the expression that shows the rate of change of the pressure as related to the rate of change of the volume.
dP/dt= ……………………
Evaluate the rate of change of the pressure at the instant when the gas fills 2〖ft〗^3, the pressure is 3000 lb/〖ft〗^2 , and the gas is expanding at 2〖ft〗^3/min
dP/dt= ……………………………
PV = k
V dP/dt + P dV/dt = 0
dP/dt = -(P/V) dV/dt
Now just plug in your numbers.
By the way, this is true only if the temperature remains constant. More generally,
PV = kT
dP/dt = (dP/dV) * (dV/dt)
Using the chain rule of differentiation, we can find the expression for dP/dt.
Given Boyle's law equation PV = k, we can rearrange it as:
P = k/V
Taking the derivative of both sides with respect to volume (V), we get:
(dP/dV) = -k/V^2
Now, let's find the rate of change of the volume (dV/dt) at the given condition.
Given: V = 2 ft^3, dV/dt = 2 ft^3/min
Substituting the given values, we have:
(dP/dt) = (-k/V^2) * (dV/dt)
= (-k/(2^2)) * (2)
= (-k/4) * 2
= -k/2
Now, let's evaluate the rate of change of the pressure at the given conditions.
Given: V = 2 ft^3, P = 3000 lb/ft^2, dV/dt = 2 ft^3/min
Substituting the given values, we have:
(dP/dt) = -k/2
To evaluate the rate of change of pressure, we need the value of the constant k. Since it is not provided, we cannot directly evaluate the exact value of (dP/dt) at the given conditions without k.
To find the expression for the rate of change of the pressure as related to the rate of change of the volume, we differentiate Boyle's law equation with respect to time:
PV = k
Differentiating with respect to time (t):
P(dV/dt) + V(dP/dt) = 0
Since we are looking for the expression dP/dt, we can isolate it:
dP/dt = - (P(dV/dt)) / V
Now let's evaluate the rate of change of the pressure using the given values:
Given:
V = 2 ft^3/min (rate of change of volume)
P = 3000 lb/ft^2 (pressure)
dV/dt = 2 ft^3/min (rate of change of volume)
Substituting the values into the expression for dP/dt:
dP/dt = - (P(dV/dt)) / V
= - (3000 lb/ft^2 * 2 ft^3/min) / 2 ft^3
= - (6000 lb*ft/min) / 2 ft^3
= - 3000 lb/min
Therefore, the rate of change of the pressure at the given instant is -3000 lb/min.
To find the expression that shows the rate of change of the pressure as related to the rate of change of the volume, we need to take the derivative of the Boyle's law equation with respect to time.
Given Boyle's law equation: PV = k
Differentiating both sides of the equation with respect to time (t), we get:
d(PV)/dt = dk/dt
Using the product rule of differentiation on the left-hand side, we have:
P * dV/dt + V * dP/dt = 0
Rearranging the equation to solve for dP/dt, we get:
dP/dt = - (P * dV/dt) / V
Now, let's evaluate the rate of change of pressure (dP/dt) at the given instant when the gas fills 2 ft^3, the pressure is 3000 lb/ft^2, and the gas is expanding at 2 ft^3/min.
We are given:
V = 2 ft^3
P = 3000 lb/ft^2
dV/dt = 2 ft^3/min
Substituting these values into the expression we derived earlier:
dP/dt = - (P * dV/dt) / V
= - (3000 lb/ft^2 * 2 ft^3/min) / 2 ft^3
= - 6000 lb/ft^2/min
Therefore, the rate of change of pressure (dP/dt) at the given instant is -6000 lb/ft^2/min.