Question

Boyle's 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 900 cm3, the pressure is 100 kPa, and the pressure is increasing at a rate of 40 kPa/min. At what rate is the volume decreasing at this instant?

Answer 1

Answer:

The value of rate of decrease of volume = - 3600 $\frac{k pa}{min}$

Step-by-step explanation:

According to Boyle's law P V = C ------- (1)

Pressure ( P ) = 100 k pa = 10 $\frac{N}{cm^{2} }$

Volume ( V ) = 900 $cm^{3}$

Put these values in equation ( 1 ) we get,

⇒ C = 10 × 900 = 9000 N-cm = 90 N-m

Differentiate Equation ( 1 ) with respect to time we get,

⇒ V $\frac{dP}{dt}$ + P $\frac{dV}{dt}$ = 0

⇒ V $\frac{dP}{dt}$ = - P $\frac{dV}{dt}$

⇒ $\frac{dV}{dt}$ = - $\frac{V}{P}$ $\frac{dP}{dt}$ ---------- (2)

This equation gives the rate of decrease of volume.

Given that Rate of increase of pressure = $\frac{dP}{dt}$ = 40 $\frac{k pa}{min}$

C = 90 N-m

P =  $10^{5}$ pa = 10 $\frac{N}{cm^{2} }$

V = 900 $cm^{3}$

Put all the above values in equation 2 we get,

⇒  $\frac{dV}{dt}$ = - $\frac{900}{10}$ × 40

⇒  $\frac{dV}{dt}$ = - 3600 $\frac{k pa}{min}$

This is the value of rate of decrease of volume.