At what rate in cubic centimeters per minute is the volume increasing at this instant?

Question

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV^(1.4)=C where C is a constant. Suppose that at a certain instant the volume is 680 cubic centimeters and the pressure is 93 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant? 

Answer 1

First calculate the value of the constant

93*680^1.4=C => C=886614. now we can use this equation:

(93-7)*(680+V)^1.4 = 886614

being V the rate of increase of volume in the same minute

(680+V)^1.4 = (886614)/(86)

taking the root 1.4 in both sides we obtain:

680 + V =735

V=rate of volume increase per minute

V= 55 cm³/min.

Answer 2

This problem can be solved by differentiation, with respect to time, of the equation

 

   P V^1.4  = C

 

The derivative of the right hand side is zero, and the derivative of the left hand side can be worked out

using the product rule and the power rule.   The resulting equation can be written in the form

 

     (1/V) dV/dt = - (1/1.4) (1/P) dP/dt

 

Writing it in this form makes it clear that both sides have units of time (minutes).  Thus we can substitute

 P = 93,  V = 680 and dP/dt = - 7  and calculate that

 

  dV/dt =  36.6 cm³/min

 

  Two additional remarks are:

Differentiating an equation will lead to another equation

involving derivatives 

 

The number 1.4 (= 7/5) is appropriate for diatomic molecules such as N2 and O2.   For atomic gases such

 as Ne and Ar, the number is 5/3  and for polyatomic gases such as methane, it is 4/3