Respuesta :
Answer:
The volume decreasing at the rate of [tex]200\:\frac{cm^3}{min}[/tex].
Step-by-step explanation:
We need to find at what rate is the volume decreasing [tex]\frac{dV}{dt}[/tex], when [tex]V=800\:cm^3[/tex], [tex]P=160\:kPa[/tex], and [tex]\frac{dP}{dt} =40 \:\frac{kPa}{min}[/tex].
We know that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation [tex]P V = C[/tex], where C is a constant.
So, we can differentiate this equation with respect the time. For this we use the Chain Rule
[tex]\frac{d}{{dx}}\left[ {f\left( u \right)} \right] = \frac{d}{{du}}\left[ {f\left( u \right)} \right]\frac{{du}}{{dx}}[/tex]
and the Product rule
[tex]\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)[/tex]
to differentiate both sides with respect to time.
[tex]V\frac{dP}{dt} +P\frac{dV}{dt} =\frac{d}{dt} C\\\\V\frac{dP}{dt} +P\frac{dV}{dt} =0[/tex]
Next, we substitute the given values
[tex]800\cdot 40+160\cdot\frac{dV}{dt} =0[/tex]
and we solve for [tex]\frac{dV}{dt}[/tex]
[tex]800\cdot 40+160\cdot\frac{dV}{dt} =0\\\\160\cdot\frac{dV}{dt} =-800\cdot 40\\\\\frac{dV}{dt} =\frac{-800\cdot 40}{160} =-200\:\frac{cm^3}{min}[/tex]
