Question

Find the rate of change of the volume when the pressure is \(20{\rm{kPa}}\) and the temperature is\(320{\rm{k}}\).

Short answer

The rate of change of volume is\(\frac{{dV}}{{dt}} = - 0.27\;{\rm{L}}/{\rm{s}}\).

Step-by-step solution

Chain rule

"Suppose that\(z = f(x,y)\)is a differentiable function of\(x\)and\(y\), where\(x = g(t)\)and\(y = h(t)\)are both differentiable functions of\(t\). Then,\(z\)is differentiable function of\(t\)and\(\frac{{dz}}{{dt}} = \frac{{dz}}{{dx}} \cdot \frac{{dx}}{{dt}} + \frac{{dz}}{{dy}} \cdot \frac{{dy}}{{dt}}\)”

Step 2: Estimate the rate of change of the volume

As given the equation is \(PV = 8.31T\) where,\(V\) is volume, \(P\) is pressure and \(T\) is temperature.

The values of \(P = 20{\rm{kPa}},T = 320\;{\rm{K}},\frac{{dP}}{{dt}} = 0.05\) and\(\frac{{dT}}{{dt}} = 0.15\)

The equation\(PV = 8.31T\), can be rewritten as \(V = 8.31\frac{T}{P}\)

By using chain rule, obtain the value of \(\frac{{dV}}{{dt}}\)

\(\begin{aligned}{l}\frac{{dV}}{{dt}} = \frac{\partial }{{\partial T}}\left( {8.31\frac{T}{P}} \right)\frac{{dT}}{{dt}} + \frac{\partial }{{\partial P}}\left( {8.31\frac{T}{P}} \right)\frac{{dP}}{{dt}}\\\frac{{dV}}{{dt}} = 8.31\left( {\frac{1}{P}} \right)\frac{\partial }{{\partial T}}(T)\frac{{dT}}{{dt}} + 8.31T\frac{\partial }{{\partial P}}\left( {\frac{1}{P}} \right)\frac{{dP}}{{dt}}\\\frac{{dV}}{{dt}} = 8.31\left( {\frac{1}{P}} \right)(1)\frac{{dT}}{{dt}} + 8.31T\left( { - {P^{ - 2}}} \right)\frac{{dP}}{{dt}}\\\frac{{dV}}{{dt}} = \frac{{8.31}}{P}\frac{{dT}}{{dt}} - 8.31\frac{T}{{{P^2}}}\frac{{dP}}{{dt}}\end{aligned}\)

Substitute \(P = 20{\rm{kPa}},T = 320\;{\rm{K}},\frac{{dP}}{{dt}} = 0.05\) and \(\frac{{dT}}{{dt}} = 0.15\) to the above equation, \(\begin{aligned}{l}\frac{{dV}}{{dt}} = \frac{{8.31}}{{20}}(0.15) - 8.31\frac{{320}}{{{{(20)}^2}}}(0.05)\\\frac{{dV}}{{dt}} = 8.31\left( {\frac{{0.15}}{{20}} - \frac{{320(0.05)}}{{400}}} \right)\\\frac{{dV}}{{dt}} = 8.31\left( {\frac{{0.15}}{{20}} - \frac{{16}}{{400}}} \right)\\\frac{{dV}}{{dt}} = 8.31\left( {\frac{{0.15}}{{20}} - \frac{1}{{25}}} \right)\end{aligned}\)

Simplify further as follows,

\(\begin{aligned}{l}\frac{{dV}}{{dt}} = 8.31\left( {\frac{{0.15}}{{20}} - \frac{1}{{25}}} \right)\\\frac{{dV}}{{dt}} = 8.31( - 0.0325)\\\frac{{dV}}{{dt}} = - 0.2701\end{aligned}\)

Hence, \(\frac{{dV}}{{dt}} = - 0.27\;{\rm{L}}/{\rm{s}}\)

Therefore, it can be concluded that the rate of change of volume is\(\frac{{dV}}{{dt}} = - 0.27\;{\rm{L}}/{\rm{s}}\).