Density of CH4
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What is the density of CH4 at 0.15 kilograms and 343 K? Is there any formula to calculate it? Thank you
For gases at low pressures, the ideal gas law can be used for calculations. That is: ρ = PM/(RT), where ρ represents density, P represents pressure, M represents the molecular weight of the gas, R represents the gas constant, and T represents temperature. This is provided for your reference.
0.0853 kg/m3. If I’m not mistaken, the density of a gas under standard conditions can be estimated using M/22.4, and then the value can be determined using the Clausius-Clapeyron equation PM=ρRT. Since the pressure is low and the temperature isn’t high, the accuracy should be fine. Last edited by cnjae on 2008-2-21 17:10.]
That can’t be right; a density of 305.3K already corresponds to 201 kg/m3
This data must be wrong too; is it that large?
That’s correct; it’s possible to make a mistake here – the value should be 1.4 kg/m3. Using PV=nRT, we get PV=(m/M)RT; then PM=(m/V)RT, which simplifies to PM=ρRT. Therefore, ρ=PM/RT. Substituting the values, we get ρ=(250000*16)/(8.314*343), resulting in ρ=1402 g/m3, or 1.4 kg/m3
I have just started learning this craft on my own, and I still face many problems. Thank you all for your help!
ρ=PM/RT is the ideal gas law, from which the formula for calculating gas density is derived. Just apply it directly.
Question: The original question states that the pressure is 0.15 kilograms; what is the relationship between this and 250,000? I believe that, in terms of gauge pressure, P = 0.15 × 98100 + 101325 = 116040; and ρ = 116.04 KPa × 16 / (8.314 × 343) = 0.651 kg/m^3
What was said on the 9th floor is correct; I misread it as 0.15 kilograms (116040 Pa) instead of 1.5 kilograms (248475 Pa ≈ 250000 Pa). This post was last edited by lsswyb01 on 2008-2-22 18:29.]
Gases at low pressure with temperatures close to room temperature can be treated as ideal gases, but they must be non-polar or weakly polar; for highly polar gases, the calculation errors can be significant.