Respuesta :
Answer:
The value of rate of decrease of volume = - 3600 [tex]\frac{k pa}{min}[/tex]
Step-by-step explanation:
According to Boyle's law P V = C ------- (1)
Pressure ( P ) = 100 k pa = 10 [tex]\frac{N}{cm^{2} }[/tex]
Volume ( V ) = 900 [tex]cm^{3}[/tex]
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 [tex]\frac{dP}{dt}[/tex] + P [tex]\frac{dV}{dt}[/tex] = 0
⇒ V [tex]\frac{dP}{dt}[/tex] = - P [tex]\frac{dV}{dt}[/tex]
⇒ [tex]\frac{dV}{dt}[/tex] = - [tex]\frac{V}{P}[/tex] [tex]\frac{dP}{dt}[/tex] ---------- (2)
This equation gives the rate of decrease of volume.
Given that Rate of increase of pressure = [tex]\frac{dP}{dt}[/tex] = 40 [tex]\frac{k pa}{min}[/tex]
C = 90 N-m
P = [tex]10^{5}[/tex] pa = 10 [tex]\frac{N}{cm^{2} }[/tex]
V = 900 [tex]cm^{3}[/tex]
Put all the above values in equation 2 we get,
⇒ [tex]\frac{dV}{dt}[/tex] = - [tex]\frac{900}{10}[/tex] × 40
⇒ [tex]\frac{dV}{dt}[/tex] = - 3600 [tex]\frac{k pa}{min}[/tex]
This is the value of rate of decrease of volume.
