Answer:
The volume increases at 35.71cm^3/min
Step-by-step explanation:
Given
[tex]PV^{1.4} = C[/tex]
[tex]V = 400cm^3[/tex]
[tex]P =80kPa[/tex]
[tex]\frac{dP}{dt} =-10kPa/min[/tex]
Required
Rate at which volume increases
[tex]PV^{1.4} = C[/tex] [tex]V = 400cm^3[/tex] [tex]P =80kPa[/tex]
Differentiate: [tex]PV^{1.4} = C[/tex]
[tex]P*\frac{dV^{1.4}}{dt} +V^{1.4}*\frac{dP}{dt} = \frac{d}{dt}C[/tex]
By differentiating C, we have:
[tex]P*\frac{dV^{1.4}}{dt} +V^{1.4}*\frac{dP}{dt} = 0[/tex]
Rewrite as:
[tex]P*(1.4)*V^{0.4}* \frac{dV}{dt} + V^{1.4}*\frac{dP}{dt} = 0[/tex]
Solve for [tex]\frac{dV}{dt}[/tex]
[tex]P*(1.4)*V^{0.4}* \frac{dV}{dt} =- V^{1.4}*\frac{dP}{dt}[/tex]
[tex]\frac{dV}{dt} =- \frac{V^{1.4}*\frac{dP}{dt} }{P*(1.4)*V^{0.4}}[/tex]
Substitute values
[tex]\frac{dV}{dt} =- \frac{400^{1.4}*-10 }{80*(1.4)*400^{0.4}}[/tex]
[tex]\frac{dV}{dt} =\frac{400*10 }{80*1.4}[/tex]
[tex]\frac{dV}{dt} =\frac{4000 }{112}[/tex]
[tex]\frac{dV}{dt} =35.71cm^3/min[/tex]
