evaluate change in pressure

Question

When something causes the volume of a gas to change very rapidly, the pressure changes also. An approximate relation between the volume and pressure of the gas can be described as, the pressure, P, varies inversely as the 3/2 power of the volume, V.     P=300 kPa when V=100 cm³  

 

a) write equation

b) derive an equation that expresses the change in pressure with respect to the change in volume.

c) evaluate the change in pressure per change in volume for V= 100 cm³

Answer 1

Ah! So this is pretty cool! Pretty much it's saying that pressure and volume are related (temperature too, but that's for another time). So let's start.

We know from Boyle's Law that P•V = k, where k is just some number. Happens a lot in math. Very fancy stuff here.

We also know that P and V are related, namely, that P • V^3/2 = k.

We know P=300 kPa and we know V = 100 cm³, so let's go ahead and plug those in now.

P • V^3/2 = (300 )(100 )^3/2 = 300,000

Right on! Now we know k! From here we can pretty much do whatever we want to make other equations. We want to get P by itself, so we just divide both sides by V^3/2 and get:

P = 300,000 / V^3/2

That's it for part a! That's the equation!

For part b, we get into some calculus. Rates of change are our derivatives, so a rate of change of pressure with respect to volume looks like:

dP/dV = ???

Now, we already have our equation for P, so we can just derive it from there to get our answer.

P = 300,000 / V^3/2

dP/dV = 300,000 (-3/2)(V^-5/2) = -450,000 / V^5/2

An easy way to think about deriving fractions, is to make them not fractions first. Power in the denominator can be come negative powers like so:

300,000 / V^3/2 = 300,000 • V^-3/2

Then, just chain rule it out. Bring the -3/2 to the front of 300,000 and subtract 1 from -3/2 to get -5/2. From there it's just multiplication and a little organization.

Lastly, we just evaluate by plugging in V = 100 cm³.

dP/dV = -450,000 / (100 )^5/2 = -450,000 / 100,000 = -4.5 kPa / cm³

As for units, just remember that the units of dy/dx - or dP/dV in our case - always come back in the end. So dP/dV would just be kPa / cm³.

That's it! Plus, remember how the question said "the pressure, P, varies inversely..."? Well, look at your answer! Volume went up, so pressure went down! There's a nice little sanity check!