Assuming ideal gas behavior, we know that
P = ÏRT
So,
Ï=P/RT
where P (atm) is pressure, Ï is molar density (mol L-1), R is a constant (0,082 atm L mol-1 K-1), and T is temperature (K)
10 bar are 9.87 atm.
Therefore, assuming ambient temperature 25°C (298 K):
Ï=(9.87)/(0,082*298) = 0.404 mol/L
Oxygen (O2) has a molecular weight of 32g/mol so:
Ï= 0.404 mol/L * 32g/mol * (1 Lt/1000 cm3) = 0.013 g/cm3
according to density=P/RT
density at 10 bar=1e6 Pa
is 11.6 Kg/M3
MolarMass = [density x gas constant x temperature(in kelvin)] / pressure (in atm)
The pressure is 20,68 at.
multiply the esoteric value by the mole fraction of the color of the moon
Density= mass/volume 25/10 = 2.5 g/cm3
As the gas warms it will want to expand and as the pressure is constant it is being allowed to do so. This means that the same number of molecules (the original mass) is/are now taking up a bigger volume and as density is a measure of mass over volume ie:- Density = Mass/volume it is obvious what the density will do because the mass is constant and the volume is increasing. Mass = 10 Volume 1 = 20 Volume 2 = 40 State 1 Mass/Volume 1 = 10/20 = density 0.5 State 2 Mass/Volume 2 = 10/40 = density 0.25
At standard temperature and pressure it is approx 1.3*10-3 grams per mL.
10 m depth is 2 bar pressure.
At these conditions, the ideal gas law should give a very good prediction of molar density of a gas. Solving the ideal gas law for molar density you get: n/V = P/RT For the stated conditions this means n/V = (1 bar)/[(8.3144622 x 10-5 m3 bar K−1 mol−1)(300 +273.15)K] = 20.98445 moles/m3 Note that it is impossible to calculate the mass density of the gas unless you specify the composition of the gas.
Air at 10 celsius and 1 bar(absolute) has density 1.25 kg/m3. By 6 bar do you mean gauge or absolute pressure? To apply a pressure factor you must use absolute, ie the pressure above a vacuum. Normal atmospheric conditions are 1 bar (abs). If you mean 6 bar(abs) the density becomes 6 x 1.25 = 7.5 kg/m3. On the other hand if you mean 6 bar (gauge) this is 7 bar (abs) and the density would be 7 x 1.25 =8.75 kg/m3. Either way, you just divide the weight of air in kg by the density to get the volume.
It means the makers claim it is waterproof to 10 bar pressure. One bar is normal atmospheric pressure, about 14.7 psi, so it should withstand 10 times normal pressure. A depth of approximately 330 feet.
90 meters. Every 10 meters, the pressure increases by approximately 1 bar, to this, you have to add the atmospheric pressure, which is also approximate 1 bar.
0.1 µPa (10-12 bar)
It means design pressure in 10 Bar
If you mean in the ocean, approximately every 10 meters pressure increases by 1 bar. Assuming you want absolute pressure, at the surface you already have a pressure of approximately 1 bar - the atmospheric pressure. You can base your calculations on that.
10 bar
PN is pressure rating of pipe. the number gives you the resistance against pressure. ın that case it is 10 bar.
P = ( h ) ( d ) ( g )P = ( 5 m ) ( 1000 kg/m^3 ) ( 9.807 m/s^2 ) = 49035 N/m^2 = 49035 PaP = ( 49035 Pa ) ( 1.01325 bar / 101325 Pa )P = 0.4904 bar