GASES
A perfect (or ideal) gas is assumed.
This is accurate enough if the pressure is not too high, nor
the gas close to liquifaction point (saturation).
Calculations with such gases obey;
1. The perfect gas law pV
= mRT
2. The 1st law of Thermo U2-U1= Q-W (everything
does!)
See table for various gases processes that determine how we
treat Q=mcT, and W=pV
SmartBoard Notes: Gases.pdf Gases.one
Perfect Gas
The perfect gas
equation;
pV = mRT
Where;
p = pressure
(Pa) absolute!
V = volume (m3)
m = mass of
gas (kg)
R = 8.314472
/ M (J/kgK) where M = relative molecular mass (no
units)
T =
temperature (K) Kelvin!
Notes:
Kelvin (K) = Celsius (oC) + 273.15
Standard Pressure is defined as one atm. or 760.0 mm Hg or 101.325 kPa.
Standard Temperature is 20oC
STP = Standard Temperature and Pressure
General
Gas Equation (Also called Combined Gas Equation)
For before-and-after
processes on a constant mass of gas, using the perfect gas pV/T = mR =
constant, so;
Summary of Special Gas Processes
Just use this table for all your gas problems. From Kinsky p113.The polytropic index varies according to the type of process, as shown in the table below.
Polytropic index |
Relation | Effects |
---|---|---|
n < 0 | — | An explosion occurs |
n = 0 | pV0
= p (constant) |
Isobaric: Equivalent to an isobaric process (constant pressure) |
n = 1 | pV = NkT (constant) |
Isothermal: Equivalent to an isothermal process (constant temperature) |
1 < n < γ | — | Polytropic: A quasi-adiabatic process such as in an internal combustion engine during expansion, or in vapor compression refrigeration during compression |
n = γ | — | Adiabatic: γ = is the adiabatic index, yielding an adiabatic process (no heat transferred) |
— | Equivalent to an isochoric process (constant volume) |
Perfect Gas Simulator
Tim Lovett 2010
See also: NASA Gas Lab Animations
The simple processes;
CONSTANT VOLUME (ISOCHORIC) |
(Gay-Lussac's Law 1809) Constant Volume Modify Temperature Watch Pressure change |
CONSTANT PRESSURE (ISOBARIC) |
(Charles' Law 1787) Constant Pressure Modify Temperature Watch Volume change |
CONSTANT TEMPERATURE (ISOTHERMAL) |
(Isothermal: Slow enough to
allow complete heat transfer) Constant Temperature Modify Volume Watch Pressure change |
Example:
This 1.25L PET (Polyethylene terephthalate) bottle was pumped up to 150 PSI.
How much energy was released when it exploded?
Since this gas process happened so quickly (faster than 1/240th of a
second), there is no time for heat transfer.
The gas process is adiabatic, so n
= γ. (Where, for air, γ
= 1.4)
Work =
We need to find the pressure and volumes in the first and second states.
p1 = 150 PSI = 1034.214 kPa =
1034214 Pa
p2 = atmospheric = 101300 Pa
V1 = 1.25 L = 0.00125 m3
V2 = ?
Use the relation:
pVn= constant.
In other words: p1V1n= p2V2n
Therefore we can find
V2;
p2V2n = p1V1n
V2n = (p1V1n) / p2
V2 = ((p1V1n) / p2)^(1/n)
= ((1034214 * 0.001251.4) / 101300) ^ (1/1.4)
= 0.006571 m3
This is the volume to which the air expanded after the bottle burst. It expands only 5.25 times, even though the pressure was 10.2 times. Had it been an isothermal process (constant temperature) it would have to draw heat in during expansion, making it expand 10.2 times. Why is this final volume less when adiabatic? Because as air expands it gets colder, which makes the volume smaller.
Now put these figures back into the Work equation:
Work = (p1V1 - p2V2) / (n-1)
= (1034214 * 0.00125 - 101300 * 0.006571) / (1.4-1)
= 1567.813 Joules (1.5 kJ)
How does this compare to the energy of a bullet?
Firearm (except listed air guns) | Caliber | Muzzle energy | ||
---|---|---|---|---|
Joules | Wh | ft-lbs | ||
air gun spring | .177 | 20 | 0.01 | 15 |
air gun mag spring | .22 | 30 | 0.01 | 22 |
air gun PCP | .22 | 40+ | 0.01 | 30+ |
pistol | .22LR | 159 | 0.04 | 117 |
pistol | 9 mm | 519 | 0.14 | 383 |
pistol | .45 ACP | 564 | 0.16 | 416 |
pistol | .357 Magnum | 873 | 0.24 | 640 |
pistol | 10mm | 1,057 | 0.29 | 775 |
1.25L PET bottle | 1 MPa | 1,500 | ||
rifle | 5.56×45 mm | 1,796 | 0.5 | 1,325 |
rifle | 7.62×39 mm | 2,070 | 0.58 | 1,527 |
rifle | 7.62 × 51 mm | 3,799 | 1.06 | 2,802 |
rifle | .338 Lapua Magnum | 6,634 | 1.84 | 4,893 |
heavy | .50 BMG | 15,037 | 4.18 | 11,091 |
heavy | 14.5 × 114 mm | 32,000 | 8.89 | 23,744 |
The PET bottle released more energy than a 10mm pistol, and a bit less than a 5.56mm rifle. Glad it didn't pop while I was holding it.
What was the temperature of the air after the explosion?
For an adiabatic expansion,
Assming T1 = 25 oC (298 K)
T2 = ( (p2 / p1) ^ (n-1/n) ) * T1
= ( (101300/1034214) ^ (1.4 - 1/1.4) ) * 298
= 60.58 K (-212.4 oC)
What was the mass of air in the pressurized PET bottle?
pV = mRT
m = pV / RT = 1034214*0.00125 / (287 * 298) = 0.01512 kg (15 grams)
Compare this to atmospheric pressure for 1.25L.
1 cubic meter of air at atmospheric pressure and 25 oC will have a mass of...
m = (101300*1) / (287 * 298) = 1.1844 kg
So for 1.25 L, the mass is: 1.1844 *1.25/1000 = 0.0014805 kg (1.5 grams)
As expected, the compressed air is 10 times the mass of atmospheric air.
Questions:
Homework Assignment: Kinksy
Do all questions; Chapter 5: Gases
5.1 to 5.25 (page 114-117) Perfect Gas Laws