2.1. Gas relative density (d) equal relative density (ρ₁ and ρ₂) gases (under the same pressure and temperature):

d = ρ₁: ρ₂ ≈ M₁ :M₂ (2.1)

where m₁ and m₂ -molecular mass of gases.

Relative density of gas:

in relation to the air: d ≈ m/29
in relation to hydrogen: d ≈ m/2

where m, 29 and 2 -the molecular mass of the gas, air and hydrogen.

2.2. Weighted number of and (in g ) gas in this volume V (in DM³):

• a = m *1.293 *g *273 * V /28.98 (273 +t) *760 = 0.01605 * r * m * V/273 + t (2.2)

where m is the molecular weight of the gas, r-gas pressure, mm Hg., t is the temperature of the gas, ⁰With.

The amount of gas in grams per 1 DM³ under normal conditions

a = 1,293 : d

where d is the relative density of gas in relation to air.

2.3. Volume V, occupied by the data sender number and strip:

V = a * 13.9 *760*(273 +t) /M * r (2.4)

2.5. Gas mixtures

Weight (in g) mix n shaped components, with volume V₁, V₂ … (V)_n and molecular mass m₁, M₂ … M_n, equal

where 22,4 -volume 1 a mole of a substance in gaseous state at 273 To and 101,32 kPa (0° C and 760 mm. RT. article)

As the volume of the mixture of V = V₁ + V₂ + ... + V_n, it 1 DM³ It has a lot of:

The average molecular mass m of gas mixture (When additivity properties) equal:

The concentration of the gas mixture components expressed most often in volume percentage. Concentration (V₁/V·100) numerically matches with a fraction of the partial pressure of component (g₁/g·100) and with its mol′noj concentration (M₁/M·100).

Shares of individual components (i) in a gas mixture is equal, %

mass volumetric

where q_i -mass content of component i in the mixture.

In equal volumes of different gases under identical conditions contain equal numbers of molecules, so

g₁:g₂: … = V₁:V₂: … = М₁:M₂:...

where m is the number of moles of.

The number of moles of a component:

etc.

If gas is in some circumstances (P, T) and it is necessary to determine its volume or mass of other conditions (R ´, T ´), use formulas:

to recalculate the amount of

for the calculation of the mass

At t = const partial pressure P_us saturated steam in a gas mixture regardless of total pressure constantly. When 101,32 KPA and t k 1 a mole of gas or vapor occupies a volume of 22,4 (T/273)DM³. If the vapor pressure at that temperature is equal to r_us, the volume of 1 Mole is equal to:

Thus, weight of 1m³ a pair of molecular mass m at a temperature t and pressure p_us equal, in the g/m³

 

 

Knowing the mass content of saturated steam at 1 m³ mix, You can calculate its pressure:

Dry gas volume calculate by the formula:

where r_{us, t} -pressure saturated steam at temperature t.

Cast dry volume V_(T,P)Suh. and humid V_(T,P)VL. gases to normal conditions (n.u.) (273 To and 101,32 kPa) produce formulas:

Formula

use to recalculate the amount of wet gas, in the r and t, other r ´, T ´, provided, that with the change of temperature changes and the equilibrium vapour pressure. The expression for the calculation of the volume of gas in various conditions similar to:

If the pressure of the water vapour saturated steam at any temperature is equal to r_us., and you must calculate G_n.. -the content of it in 1 m³ gas n.a., use the equation (1.2), but in this case there is no saturation temperature t, and equal 273 To.

It follows, what:

G_n.. = 4,396·10^-7 Mr_us..

Pressure saturated steam, If you know its contents in 1 m³ When Mr.. calculate using the formula: