The Question

The pressure, volume, and temperature of an ideal gas are related by the equation PV=8.31T (when pressure is measured in kilopascals). Find the rate at which the pressure is changing when the temperature is 300 K and increasing at a rate of 0.1 K s^{-1}, and the volume is 100 L and increasing at a rate of 0.2 L s^{-1}.

Solution

First of all, solving for P:

P(T,V)=\frac{8.31T}{V}.

Now we can go further and say that both T and V are functions of time, t. So we have:

P(T(t),V(t))=\frac{8.31T(t)}{V(t)}=8.31T(t)[V(t)]^{-1}.

Now changes to pressure, \Delta P are going to result from independent changes to temperature, \Delta T, and pressure \Delta V.

\Delta P =\text{changes due to }\Delta T+\text{changes due to }\Delta V.

Now changes to temperature and volume are due to changes in time, \Delta t. It can be shown that this change manifests itself as (with the obvious generalisation to derivatives of functions of the form z=f(x(t),y(t))):

\frac{dP}{dt}=\frac{\partial P}{\partial T}\frac{dT}{dt}+\frac{\partial P}{\partial V}\frac{dV}{dt}.

(See: http://library.ucc.ie/search~S0?/Xcalculus%20by%20stewart&SORT=D/Xcalculus%20by%20stewart&SORT=D&SUBKEY=calculus%20by%20stewart/1,3576,3576,B/frameset&FF=Xcalculus%20by%20stewart&SORT=D&3,3, for details of this.)

Now to calculate this rate of change at (T,V)=(300,100) we must put these values into P_T, P_V, so:

\frac{dP}{dt}=\left.\frac{\partial P}{\partial T}\right|_{(T,V)=(300,100)}\frac{dT}{dt}+\left.\frac{\partial P}{\partial V}\right|_{(T,V)=(300,100)}\frac{dV}{dt}.

Since we know each of the constituents of this equation (dT/dt=+0.1, dP/dV=+0.2) all that is left is to do the calculations.

\frac{\partial P}{\partial T}=8.31[V(t)]^{-1}=8.31(100)^{-1} at (300,100).

\frac{\partial P}{\partial V}=8.31T(t)(-[V(t)]^2)=-8.31(300)(100)^{-2}.

Hence:

\frac{dP}{dt}=(8.31(100)^{-1})\times (+0.1)+(-8.31(300(100)^{-2}))\times(+0.2)

=-0.04155 kPa.

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