How do you solve this three part calculus problem?

Question

Suppose a quantity of gas (with a fixed mass) is confined in a container of volume V. Boyle's law states that, so long as the temperature of a gas is constant, the product of the pressure of P and the volume V is a constant. Suppose that, for a certain gas, PV=850, where P is measured in lbs/sq inches and V is measured in cubic inches.

 

A. Find the average rate of change of P (with respect to V) as V increases from 200 to 300 cubic inches.

 

B. Suppose that the volume is increasing at 100 cubic inches per hour. Find the instantaneous rate of change of P with respect to time when V=200 cubic inches.

 

C. Suppose that the temperature T is allowed to change as well. The expression that relates P, T, and V is PV/T = constant. How does P change (increase, decrease or stay same) if V is fixed and T increases?

Answer 1

A. I think you're missing a piece of information, specifically time.  That said, we can calculate P at each condition of V.  If PV = 850, then P = 850/V.  So at the start, P = 850/200 = 4.25.  At the end, P = 850/300 = 2.83.  To find the average rate, we'd take the change divided by the time: (2.83 - 4.25)/t.

 

B. If we apply implicit differentiation (with a product rule), we get

 

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

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

 

When V = 200, P = 4.25, so:

 

dP/dt = -(4.25)/200 * 100 = -2.13

 

C.  If we have PV/T = C, with C a constant and V constant, then PV = CT.  We see that P and T are directly related.  Therefore, if T increases (with V constant), then P must increases.