# 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 $z=f(x(t),y(t))$): $\frac{dP}{dt}=\frac{\partial P}{\partial T}\frac{dT}{dt}+\frac{\partial P}{\partial V}\frac{dV}{dt}$.

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.