Boyles Law?

Question

According to Boyle's Law, when a sample of gas is compressed at a constant temperature, the pressure and volumn satisfy the equation P V = C, where C is a constant. Assume that, at a certain instant, a sample has a volume of 950 cm3, is a pressure of 120 kPa, and the pressure is increasing at a rate of 19 kPa/min. At what rate is the volume decreasing at this instant?

Answer 1

PV = C
C = 950 x 120
C = 114000

V = C/P = CP^(-1)
dV/dP = - C / P ²

dV/dt = (dV/dP) (dP/dt)
dv/dt = (- C / P ²) (19)
dv/dt = (- 114000) (19) / (120 ²)
dv/dt = - 150.4 cm ³ / min
Volume is decreasing at a rate of 150.4 cm ³ / min

Answer 2

Use implicit differentiation with respect to some "time" variable t.

(dP/dt)V + P(dV/dt) = dC/dt
(19)(950) + (120)(dV/dt) = 0
120(dV/dt) = -(19)(950) = -18050
dV/dt = -18050/120 = -1805/12

That tell us that the volume is decreasing at a rate of (-1805/12), where the units are ((cm^3)/min).

Answer 3

PV=c
The trick is to differentiate this with respect to time using the product rule because both P and V are functions of time.

P dV/dt + VdP/dt = 0
We know the values of P, V and dP/dt
plug in those values to get dV/dt.
the value is obviously negative because the volume is decreasing with time.