boyles law states that when a sample of gas is compressed at a constant temperature, the pressure ( p ) and…

boyles law states that when a sample of gas is compressed at a constant temperature, the pressure ( p ) and volume ( v ) satisfy the equation ( pv = c ), where ( c ) is a constant. suppose that at a certain instant the volume is ( 100 mathrm{~cm}^{3} ), the pressure is ( 200 mathrm{~kpa} ), and the pressure is increasing at a rate of ( 40 mathrm{~kpa} / mathrm{min} ). at what rate (in ( mathrm{cm}^{3} / mathrm{min} )) is the volume decreasing at this instant? ( mathrm{cm}^{3} / mathrm{min} )

boyles law states that when a sample of gas is compressed at a constant temperature, the pressure ( p ) and volume ( v ) satisfy the equation ( pv = c ), where ( c ) is a constant. suppose that at a certain instant the volume is ( 100 mathrm{~cm}^{3} ), the pressure is ( 200 mathrm{~kpa} ), and the pressure is increasing at a rate of ( 40 mathrm{~kpa} / mathrm{min} ). at what rate (in ( mathrm{cm}^{3} / mathrm{min} )) is the volume decreasing at this instant? ( mathrm{cm}^{3} / mathrm{min} )

Answer

Explanation:

Step1: Find the value of constant (C)

Given (P = 200) kPa and (V=100) (cm^{3}), from (PV = C), we have (C=P\times V). Substitute the values: (C = 200\times100=20000). So the equation is (PV = 20000), and we can express (V=\frac{20000}{P}).

Step2: Differentiate (V) with respect to time (t)

Using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = 20000) ((u^\prime=0)) and (v = P) ((v^\prime=\frac{dP}{dt})). (\frac{dV}{dt}=\frac{0\times P-20000\times\frac{dP}{dt}}{P^{2}}=-\frac{20000}{P^{2}}\frac{dP}{dt}).

Step3: Substitute the given values of (P) and (\frac{dP}{dt})

We know that (P = 200) kPa and (\frac{dP}{dt}=40) kPa/min. (\frac{dV}{dt}=-\frac{20000}{200^{2}}\times40). First, calculate (200^{2}=40000). Then (\frac{20000}{40000}=\frac{1}{2}). (\frac{dV}{dt}=-\frac{1}{2}\times40=- 20) (cm^{3}/min).

Answer:

(20)