Given
Pressure (P)-Volume (V) relation : #PV^1.4=C#,where C= constant
Differentiating the given equation w.r.t time (t) we get
#Pd/(dt)(V^1.4)+V^1.4(dP)/(dt)=d/(dt)(C)=0#
#=>Pxx1.4xxV^(1.4-1)xx(dV)/(dt)+V^1.4(dP)/(dt)=0#
#=>Pxx1.4xxV^(0.4)xx(dV)/(dt)+V^1.4(dP)/(dt)=0#
#=>(dV)/(dt)=-(V^1.4(dP)/(dt))/(Pxx1.4xxV^(0.4))#
#=>(dV)/(dt)=-(V(dP)/(dt))/(1.4xxP)........(1)#
Now we are also given
#" Pressure "P=93kPa#
#"Volume "V=610cm^3#
#"Rate of change of Pressure with time "(dP)/(dt)=-9" kPa/min"#
(-ve sign represents the decreasing rate with time)
Inserting these values in equation (1) we get the rate increasing volume with time #[(dV)/(dt)]#
#(dV)/(dt)=-(V(dP)/(dt))/(1.4xxP)#
#=>(dV)/(dt)=-(610cm^3xx(-9" kPa/min"))/(1.4xx93kPa)#
#=>(dV)/(dt)~~42.16" cm^3/min"#
