The pressure P in kilopascals
and volume V in liters, and temperature T in kelvins
of a mole of an ideal gas are related
by the equation PV = 8.31 times T.
We're asked to find the rate at which the volume is changing
when the temperature is 315 kelvins
and increasing at a rate of 0.05 kelvins per second
and the pressure is 17 kilopascals
and increasing at a rate of 0.02 kilopascals per second.
Let's begin by stating all the given information.
We know the temperature T equals 315 kelvins.
We know the temperature is increasing
at a rate of 0.05 kelvins per second,
which means a derivative of big T
with respect to time t is equal
to positive 0.05 kelvins per second.
We know the pressure is 17 kilopascals.
So P equals 17 kilopascals.
And finally the pressure is increasing
at a rate of 0.02 kilopascals,
which means dP/dT equals
0.02 kilopascals per second.
So all this information is given
and now we're trying to find the rate
at which the volume is changing,
and we're given the equation PV equals 8.31 times T.
So because we're looking for the rate
at which the volume is changing,
we're going to determine dV/dt.
So let's first solve this equation for V
by dividing both sides by P.
So we can say that V equals 8.31
times T divided by P--
Well if we want 8.31T times P
raised to the power of negative one.
And because V is a function of big T and P,
we'll have to apply the chain rule given here below.
So let's go ahead and write out
the chain rule using our variables.
So we'll have the derivative of V with respect to T
is equal to the partial derivative of V
with respect to big T
times the derivative of big T
with respect to time, or t,
plus the partial derivative of V with respect to P
times dP/dt.
Now let's find our partial derivatives.
So we have dV/dt equals
the partial derivative of V
with respect to t, so a different shade with respect to T,
treating P as a constant,
which would just give us 8.31 times P
raised to the power of negative one,
because the derivative of T with respect to t
would just be one.
So we have 8.31 times P to the negative 1
times we have dT/dt,
which we know the value of,
but we'll substitute that value in in the next step.
And then we have plus the partial derivative of V
with respect to P, so we'll differentiate with respect to P,
treating T as a constant,
so we're going to multiply by negative one,
and then subtract one from the exponent on P.
So we'll have negative 8.31T times P
raised to the power of negative two.
So we have negative 8.31 times T
times P raised to the power of negative two
times dP/dT.
Notice how here we can write this as a difference
rather than plus a negative.
Let's continue on the next slide.
Before we perform substitution,
let's rewrite these negative exponents
as positive exponents.
So we'd have dV/dt equals--
Here we'd have 8.31 divided by P to the first, or just P,
times dT/dt, minus 8.31T
divided by P squared
times dP/dt.
And now we'll substitute the given values in for T,
dT/dt, P, and dP/dt.
So we have 8.31 divided by P, which is 17,
times dT/dt, which we know is 0.05.
And then we have minus 8.31 times T, which is 315,
divided by P squared, which'd be 17 squared
times dP/dt, which is 0.02.
To save some time, I've already determined these values.
This value here is approximately 0.02444
minus this value here is approximately 0.18115.
So if you find this difference
and round to four decimal places,
we would have dV/dt is approximately negative 0.1567.
Now this is measuring the change in the volume
with respect to time.
And going back to the question just for a moment,
notice how the volume is in liters,
and our time is in seconds.
Which means the units here would be liters per second.
So the volume is decreasing at 0.1567 liters per second
when we have all of this given information.
I hope you found this helpful.