A cylinder of compressed air had a gauge pressure of 710 kPa

A cylinder of compressed air had a gauge pressure of 710 kPa at the start of an operation. After the operation the gauge pressure was 35 kPa. If the cylinder has a volume of 0.05 m^3, what volume of air at atmospheric pressure was delivered? The operation was slow enough so that the temperature can be assumed constant.

Solution

The amount of air in the system originally must be equal to the amount of air

remaining in the cylinder plus the amount of air that was released. Instead of dealing with

masses, we can find (1) the equivalent volume of air originally in the container if it was held

at atmospheric pressure (V1;atm), and (2) the equivalent volume of air left in the container

if it was held at atmospheric pressure (V2;atm). The difference between these two volumes

will be the amount of air released from 1 to 2.

We can find the equivalent volumes using Boyle\'s law, and since the entire system is

assumed to have the same temperature, these calculations are valid. Boyle\'s law is:

PaVa=PcVc                        (1)

where

Pa and Va are the pressure and volume, in this case, in the atmpsohere, and Pc and Vc are the pressure and volume in the compressed state.

You can always arrive at Boyle\'s Law by using the ideal gas law,

PV= mR where P is pressure, V is volume, m is mass,R is the specic mass constant and T is temper-ature. Since

mRT is constant (neither mass,R or temperature are changing here),PV remains constant in this calculation. Therefore, (PV)a = (PV)c

.Let\'s first find the equivalent volume of the original cylinder.

(2) P1V1=PatmV1;atm

Rearranging for V1;atm

(3)V1;atm=P1V1/Patm

When using Boyle\'s law or any other equation that involves mathematical operations of

pressures, such as multiplication/division in this case, the pressures need to be absolute, not gauge Same goes for operations of temperature and the use of Kelvin over Celsius.

Atmospheric pressure is 101.325 kPa.

(4) P1=P1;gauge+Patm= 710 kPa + 101.325 kPa = 811.325 kPa

This is excessive signicant gures, but we will leave it until the end.

Substituting into 3

(5)V1;atm=P1V1/Patm=811.325 kPa.0.05m³/101.325kpa =0.4m³

Because the kPa\'s cancel, we don\'t have to worry about putting the units in Pascals. As

long as the units are consistent on top and bottom, the calculation is ne. Let\'s do the

same for the remaining volume of air

(6) P2=P2;gauge+Patm= 35 kPa + 101.325 kPa = 136.325 kPa

Substituting into the equivalent of equation 3

(7)V2;atm=P2V2/Patm=34.325 kPa.0.05m³/101.325kpa =0.067m³

the difference between these two is the volume of air released at atmospheric pressure.

Vreleased;atm=V1;atm-V2;atm=0.4-0.067=0.333m³