Answer:
26.466cm³/min
Explanation:
Given:
Volume 'V'= 320cm³
P= 95kPa
dP/dt = -11 kPa/minute
pressure P and volume V are related by the equation
P[tex]V^{1.4}[/tex]=C
we need to find dV/dt, so we will differentiate the above equation
[tex]V^{1.4} \frac{dP}{dt} + P\frac{d[V^{1.4} ]}{dt} = \frac{d[C]}{dt}[/tex]
[tex]\frac{dP}{dt} V^{1.4} + P(1.4)V^{0.4} \frac{dV}{dt} =0[/tex]
lets solve for dV/dt, we will have
[tex]\frac{dV}{dt} =\frac{-\frac{dP}{dt} V^{1.4} }{P(1.4)V^{0.4} } \\\frac{dV}{dt} =- \frac{-\frac{dP}{dt} V}{P(1.4)}[/tex]
[tex]\frac{dV}{dt} = -\frac{(-11 ) 320}{95(1.4)}[/tex] (plugged in all the values at the instant)
[tex]\frac{dV}{dt}[/tex] = 26.466
Therefore, the volume increasing at the rate of 26.466cm³/min at this instant
Answer:
26.47 cm³/minute
Explanation:
We first differentiate the expression to obtain the rate of change of volume with time dV/dt
[tex]PV^{1.4} = C\\ d(PV^{1.4})/dt = dC/dt\\1.4PV^{0.4}\frac{dV}{dt} + V^{1.4}\frac{dP}{dt} = 0\\1.4PV^{0.4}\frac{dV}{dt} = -V^{1.4}\frac{dP}{dt}\\ \frac{dV}{dt} = -\frac{V}{1.4P} \frac{dP}{dt}[/tex]
Now when V = 320 cm, P = 95 kPa and dP/dt = -11 kPa/minute since it is decreasing.
We substitute these values into the expression for dV/dt. So,
[tex]\frac{dV}{dt} = -\frac{V}{1.4P} \frac{dP}{dt} = -\frac{320 cm^{3} }{1.4 X 95 kPa} X- 11 kPa/minute\\= 26.47 cm^{3}/minute[/tex]
