Question

How to calculate average pressure as volume changes via intergration?

A quantity of gas expands according to the law pv^0.9=300, where v m^3 is the volume of the gas and p N/m^2 is the pressure
a) what is the average pressure as the volume changes from 1/2 m^3 to 1 m^3?

Answer 1

`p_("ave") = 401.802 " Pa"`

Explanation:

The idea is the simple geometric one that:

`p_("ave") * (V_2 - V_2) = int_(V_1)^(V_2) \ p(V) \ dV`

`implies p_("ave") = (300 int_(1/2)^(1) V^(-0.9)\ dV)/(1 - 1/2)`

`= 6000 ( V^(0.1) )_(1/2)^(1)`

`implies p_("ave") = 401.802 " Pa"`

You would expect pressure to drop over the expansion, so reality check:

• `p_1 approx 560 " Pa"`

• `p_2 = 300 " Pa"`

• `p_2 lt p_("ave") lt p_1`, which is expected