Pressure (p) and volume (v) given in eqn #pv^1.4=k# where k is constant. At certain instant pressure is #(25gm)/(cm^3# and volume is #32cm^3#. If volume is increasing at rate of #(5cm^3)/s#, how do you find the rate at which pressure is changing?

1 Answer
Jun 10, 2018

#-5.468# grams #cms^2 sec^-1#

Explanation:

Assuming that the pressure should be in grams #cms^2# and not #cms^3# as in the question.

It is given that #pv^1.4=K# and we are told that at time # t #, Pressure # P= 25# and at the same time #t#, Volume= #32#, from this information we can work out the constant #K#.

Thus #25[32^1.4]=K#, i.e #K = 3200#

So, at time # t # , #[pv^1.4=3200]#.........#[1]#, differentiating ....#[1]# implicitly with respect to pressure and time using the product rule, in this case [ #d/dt[pv]=v[dp]/[dt]+p[dv]/[dt]#], where #p# and #v# and are both functions of # t #.

#v^1.4[dp]/[dt] + p[1.4v^[0.4]][dv]/[dt]=0#.........#[2]#. We know #[dv]/[dt] # at time #t# is #5# and we also know that at time #t#, # p #=#25# and that the volume #v =32# so we can substitute all these values into.....#[2]# and solve for #[dp]/[dt]#

#[dp]/[dt] = -[[25][1.4][32^0.4][5]/[32^1.4]# This is the rate of change of pressure at time #t#. and is equal to -#5.468 cms^2 sec^-1#. Hope this helps.