Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 900 cm3, the pressure is 200 kPa, and the pressure is increasing at a rate of 20 kPa/min. At what rate is the volume decreasing at this instant?

Respuesta :

Answer:

-90cm³/min

Explanation:

The equation is:

[tex] P*V = C [/tex]  (1)

Where P: is the pressure, V: is the volume, and C: is a constant.

To find the rate at which the volume decrease, we need to derivate equation (1):

[tex] P\cdot \frac{dV}{dt} + \frac{dP}{dt}*V = \frac{dC}{dt} [/tex]   (2)

From equation (2) we have:

P = 200 kPa

V = 900 cm³

[tex] \frac{dP}{dt} = 20 kPa/min [/tex]  

[tex] \frac{dC}{dt} = 0 [/tex]

Hence, by entering the values above in equation (2), we have:

[tex]200 kPa\cdot \frac{dV}{dt} + 20 kPa/min* 900 cm^{3} = 0 [/tex]          

[tex]\frac{dV}{dt} = - \frac{20 kPa/min* 900 cm^{3}}{200 kPa} = - 90 cm^{3}/min[/tex]          

Therefore, the rate at which the volume is decreasing is -90cm³/min.

I hope it helps you!