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
You can use the ideal gas law to find the density of oxygen at 1.00 bar and 10 degrees C. First, calculate the molar volume of gas using the ideal gas law. Then, divide the molar mass of oxygen by the molar volume to find the density.
The molar mass of air can be calculated by adding the weighted average of nitrogen, oxygen, and argon based on their mole fractions. Then, the density of air at standard temperature and pressure can be calculated using the ideal gas law. The standard molar volume of ideal gas is 22.4 L/mol at standard temperature and pressure.
For cutting 12.7mm sheet with a positive pressure torch, oxygen pressure should typically be set around 30-40 PSI, while acetylene pressure should be set around 5-10 PSI. These pressures may need to be adjusted based on the specific torch and cutting conditions. Always refer to the manufacturer's guidelines for the most accurate pressure recommendations.
The density of chlorine gas at 7.50 × 10^2 torr and 25.0ºC can be calculated using the ideal gas law. First, convert the pressure to atm (7.50 × 10^2 torr = 0.988 atm). Then, use the ideal gas law equation: PV = nRT and rearrange it to solve for density (density = PM/RT where M is the molar mass of chlorine gas). Substituting the values and calculating will give the density in g/L.
Using the ideal gas law (PV = nRT), we can solve for pressure. Given: T = 42 deg C = 315 K, V = 10 L, n = 8 moles, R = 0.0821 L.atm/mol.K. Substituting these values: P = (8 moles)(0.0821 L.atm/mol.K)(315 K) / 10 L = 20.6 atm.
You can use the ideal gas law to find the density of oxygen at 1.00 bar and 10 degrees C. First, calculate the molar volume of gas using the ideal gas law. Then, divide the molar mass of oxygen by the molar volume to find the density.
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.
Water pressure increases by approximately 1 bar for every 10 meters of depth in freshwater. At a depth of 10 meters, the water pressure would be about 1 bar, in addition to the atmospheric pressure at the surface, which is roughly 1 bar as well. Therefore, the total pressure at 10 meters depth would be about 2 bars.
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.
The density of a substance is equal to its mass divided by its volume. To calculate the density of oxygen in this case, we need to divide its mass by its volume. So, the density of 10 liters of oxygen with a mass of 0.014 would be 0.0014 g/mL.
It means design pressure in 10 Bar
0.1 µPa (10-12 bar)
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.
In water, every 10 meters you go down, the pressure increases by 1 bar, approximately. To this you must add the air pressure, which is also approximately 1 bar (depending on whether you want gauge pressure or absolute pressure).