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# Help with physics/calculus problem on Boyle's law!?

Boyle's law says PV = C. At a certain instant the volume is 200 ccs, pressure is 80 kPa, & pressure is increasing at rate 32 kPa/min.

At what rate is the volume decreasing at this instant?

### 2 Answers

- gintableLv 71 decade agoFavorite Answer
Boyle's law is only applicable for a system of an ideal gas at constant temperature, just so you know.

It is stated as:

P*V = C

Take the derivative using implicit differentiation relative to P.

d/dt (P*V) = d/dt (C)

the derivative of a constant is zero, thus:

d/dt (P*V) = 0

Use the product rule:

P*dV/dt + V*dP/dt = 0

Solve for dV/dt:

dV/dt = -V/P * dP/dt

Given:

P:=80 kPa; dP/dt:=32 kPa/min; V:=200 cm^3;

Result:

dV/dt = -80 cm^3/min

The negative sign means that volume is decreasing, just as the question statement predicted. This means volume is decreasing at 80 cubic centimeters/minute.

- nedeauLv 44 years ago
For all Charles' regulation issues, the worry-unfastened equation is: V1/T1 = V2/T2. One important factor to bear in innovations for ALL gas regulation issues is that the temperature could desire to be in Kelvin (no longer celcius). a million. 10.0 C = 283 ok. So V1/T1 = V2/T2 733/283 = 950/T2 T2 = 367 ok 2. try this one precisely an identical way: V1/T1 = V2/T2 560/285 = 25.0 / T2 T2 = 12.7 ok For Boyle's regulation, the equation is P1V1 = P2V2 a million. 622kPa (233 mL) = 988 KPa (V2) V2 = 147 mL the subsequent 2 are performed precisely an identical way, different than that for the time of #3, you're fixing for P2 instead of V2. In those issues, it extremely does not rely what the gadgets of P and V are, as long as you're consistent interior of a situation. in case you instructor says the choose arises exhibit volumes in Liters, or stress in kPa, then convert them. you are able to the two convert the respond or convert the numbers in the topics...that's not appropriate...