Question

A vertical piston cylinder device contains a gas at an unknown pressure. If the outside pressure is 100kPa, determine a) the pressure of the gas if the piston has an area of 0.2m2. And a mass of 20 kg. Assume g=9.81m/s2. b) What would the pressure be if the orientation of the device were changed and it were now upside down?

Answer 1

a)
`P_2 = 100000Pa + \frac{196N} {0.2 m^2} =100980 Pa= 100.98 Kpa \approx 101 Kpa`

b)
`P = 100000Pa -(196 N)/(0.2 m^2)=99020Pa =99.02 Kpa \approx 99Kpa`

Step-by-step explanation:

Let's consider the piston on the figure attached.

We know that the atmospheric pressure on this case is
`P_(atm)= 100 Kpa = 100000 Pa`

We know that the Are of the piston is
`A = 0.2 m^2` and we have also the mass
`m = 20Kg`

Part a

We want to find the pressure at the gas, so that means point 2, and we assume that the pressure on any point of the gas is constant and the same. So then we can find the pressure at point 2 like this:

`P_2 = P_(atm) + (F)/(A)`

And the force on this case is the weigth
`F = mg = 20 kg *9.8 (m)/(s^2)=196 N`

And then we have:

`P_2 = 100000Pa + \frac{196N} {0.2 m^2} =100980 Pa= 100.98 Kpa \approx 101 Kpa`

Part b

For this case we want to find the pressure if the device were changed and now would be upside down. On this case we will have lower pressure than from part a and would be given by:

`P = 100000Pa -(196 N)/(0.2 m^2)=99020Pa =99.02 Kpa \approx 99Kpa`