Question

When air expands adiabatically (without gaining or losing heat), its pressure \(P\) and volume \(V\) are related by the equation \(P{V^{1.4}} = C\), where C is a constant. Suppose that at a certain instant the volume is \(400\;{\rm{c}}{{\rm{m}}^3}\)and the pressure is 80 kPa and is decreasing at a rate of 10 kPa/min. At what rate is the volume increasing at this instant?

Short answer

The Volume increases at rate of \(36\;{\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/min}}\).

Step-by-step solution

Write the given conditions

From the given condition, we have \(P = 80\), \(\frac{{dP}}{{dt}} = - 10\) and \(V = 400\;{\rm{c}}{{\rm{m}}^{\rm{3}}}\).

Apply chain rule for differentiation

Chain rule of differentiation is \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}.\frac{{du}}{{dx}}\)

Apply chain rule in the given equation and simplify.

\(\begin{aligned}\frac{{dP}}{{dt}}{\left( V \right)^{1.4}} + P\left( {\frac{{dV}}{{dt}}} \right){V^{0.4}}\left( {1.4} \right) &= 0\\\frac{{dV}}{{dt}}& = - \frac{{{V^{1.4}}}}{{P \cdot \left( {1.4} \right){V^{0.4}}}}\frac{{dP}}{{dt}}\\\frac{{dV}}{{dt}} &= - \frac{{{V^1}}}{{P \cdot \left( {1.4} \right)}}\frac{{dP}}{{dt}}\\\frac{{dV}}{{dt}} &= - \frac{{400}}{{80 \cdot \left( {1.4} \right)}}\left( { - 10} \right)\\\frac{{dV}}{{dt}} &= \frac{{250}}{7}\\\frac{{dV}}{{dt}} \approx 36\end{aligned}\)

The Volume increases at rate of \(36\;{\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/min}}\).