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 610 cubic centimeters and the pressure is 89 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?
Please go back to each of your posts and let the tutors know what you do and don't understand about each one.
To expect someone here to do all your work for you is unrealistic!
i am slowly doing all the work, I just want them to verify my answers, if I have a question about a problem i will post it
48.95666
The way you typed it , I read that as
(P)(V)(1.4) = C
in that case, I did not get you answer.
If you meant (PV)^1.4 = c
or P V^1.4 = c
then we have another story
what is it?
it was ^1.4, I have it solved now however.
To find the rate at which the volume is increasing at a certain instant, we need to differentiate the given equation with respect to time.
Let's start by differentiating the equation PV^1.4 = C with respect to time (t):
(d/dt)(PV^1.4) = (d/dt)(C)
Using the product rule of differentiation, we have:
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 can assume that dP/dt = 0, as there is no heat transfer occurring.
Therefore, the equation simplifies to:
P * (dV/dt)(1.4V^0.4) = 0
Now we can substitute the values given in the problem:
P = 89 kPa
V = 610 cm^3
dP/dt = -10 kPa/minute (negative sign indicates decreasing pressure)
We are solving for dV/dt (rate at which volume is increasing).
Plugging these values into the equation, we have:
89 * (-10) * (1.4 * 610^0.4) = dV/dt
Simplifying this equation will give us the rate at which the volume is increasing at this instant.