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I am having trouble with physical applications of calculus. Can someone help me with this question?
When a gas expands without a change in temperature, the pressure 'p' and volume 'v' are given by the relation pv^1.4=k, where k is a constant. At a certain instant, the pressure is 25gm/cm^2 and the volume is 32cm^3. If the volume is increasing at a rate of 5cm^3/s, at what rate is the pressure changing at this instant?
2 Answers
- 7 years ago
log get a equation
lgP+1.4lgV=lgK
then; lgP= (-1.4)lgV+lgK this is equal to Y=mX+c
m=(-1.4), c=lgK
therefore pressure changing rate =negative 1.4
- JaredLv 77 years ago
pv^1.4 = k
--> differentiate implicitly
v^1.4 * dp/dt + 1.4pv^0.4 * dv/dt = 0
-->
dp/dt = -1.4pv^0.4 / v^1.4 * dv/dt = -1.4p/v * dv/dt
-->
dp/dt = -1.4 * 25 / (32) * 5 = -5.46875 gm/cm²
