What is the partial pressure of oxygen in atmospheric air at sea level?

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The zirconium dioxide O2 sensor working principle measures partial pressure of oxygen in a mixture of gases. This typically causes confusion amongst PST’s customers as most oxygen sensors on the market measure oxygen concentration.

But, what is partial pressure? It is a question we are asked frequently when it comes to the O2 sensor working principle. In this article, we will address the definition of partial pressure, the physics behind it, how you calculate partial pressure and how to convert the oxygen partial pressure into volumetric content for those interested in oxygen concentration.

Partial Pressure: The Definition

The partial pressure is defined as the pressure of a single gas component in a mixture of gases. It corresponds to the total pressure which the single gas component would exert if it alone occupied the whole volume.

Daltons Law: The Physics

The theory of the o2 sensor working principle is detailed here. The total pressure (Ptotal) of a mixture of ideal gases is equal to the sum of the partial pressures (Pi) of the individual gases in that mixture.

From Equation 1 it can be derived that the ratio of the number of particles (ni) of an individual gas component to the total number of particles (ntotal) of the gas mixture equals the ratio of the partial pressure (Pi) of the individual gas component to the total pressure (Ptotal) of the gas mixture.

ni	Number of particles in gas
ntotal	Total number of particles
pi	Partial pressure of gas i
Ptotal	Total pressure

Figure 1 Ptotal = P1 + P2 + P3 (Constant Volume & Temperature)

Example 1:

The atmospheric pressure at sea level (under standard atmospheric conditions) is 1013.25mbar. Here, the main components of dry air are nitrogen (78.08% Vol.), oxygen (20.95% Vol.), argon (0.93% Vol.) and carbon dioxide (0.040% Vol.). The volumetric content (%) can be equated to the number of particles (n) since the above gases can be approximated as ideal gases.

Equation 2 can be solved for the partial pressure of an individual gas (i) to get:

The oxygen partial pressure then equates to:

Pi = x 1013.25mbar = 212.28mbar

Figure 2 Partial Pressure at 0% Humidity

Of course, this value is only relevant when the atmosphere is dry (0% humidity). If moisture is present a proportion of the total pressure is taken up by water vapour pressure. Therefore, the partial oxygen pressure (ppO₂) can be calculated more accurately when relative humidity and ambient temperature are measured along with the total barometric pressure.

Firstly, water vapour pressure is calculated:

WVP = ( ) x WVPmax

WVP	Water Vapour Pressure (mbar)
HRel	Relative Humidity (%)
WVPmax	Maximum Water Vapour Pressure (mbar)

For a known ambient temperature, maximum water vapour pressure (WVPmax) can be determined from the lookup table below. The maximum water vapour pressure is also referred to as the dew point. Warmer air can hold more water vapour and so has a higher WVPmax.

Temperature (°C)	Max water vapour pressure (mbar)	Temperature (°C)	Max water vapour pressure (mbar)
0	6.1	31	44.92
1	6.57	32	47.54
2	7.06	33	50.3
3	7.58	34	53.19
4	8.13	35	56.23
5	8.72	36	59.42
6	9.35	37	62.76
7	10.01	38	66.27
8	10.72	39	69.93
9	11.47	40	73.77
10	12.27	42.5	84.19
11	13.12	45	95.85
12	14.02	47.5	108.86
13	14.97	50	123.86
14	15.98	52.5	139.5
15	17.04	55	457.42
16	18.17	57.5	177.25
17	19.37	60	199.17
18	20.63	62.5	223.36
19	21.96	65	250.01
20	23.37	67.5	279.31
21	24.86	70	311.48
22	26.43	75	385.21
23	28.11	80	473.3
24	29.82	85	577.69
25	31.66	90	700.73
26	33.6	95	844.98
27	35.64	100	1013.17
28	37.78	110	1433.61
29	40.04	120	1988.84
30	42.42	130	2709.58

Partial oxygen pressure then equates to:

ppO2 = (BP - WVP) x ( )

ppO2	Partial Pressure O2 (mbar)
BP	Barometric Pressure (mbar)
WVP	Water Vapour Pressure (mbar)

Example 2 below describes the effect of humidity reducing the partial oxygen pressure and therefore the volumetric content of oxygen.

Example 2:

On a typical day, the following information is recorded from a calibrated weather station:

Temperature	22°C
Humidity	32%
Barometric Pressure	986mbar

Using the Water Vapour Pressure look up table above, WVPmax = 26.43mbar.

WVP = ( ) x 26.43 = 8.458mbar

Partial oxygen pressure then equates to:

ppO2 = (986 - 8.458) x ( ) = 204.795mbar

As we now know the oxygen partial pressure and the total barometric pressure we can work out the volumetric content of oxygen.

O2% = ( ) x 100 = 20.77%

Gas pressures in the atmosphere and body determine gas exchange: both O2 and CO2 will flow from areas of high to low pressure.

Learning Objectives

Describe how gas pressure influences the flow of gases during respiration

The respiratory process can be better understood by examining the properties of gases. Gases move freely with their movement resulting in the constant hitting of particles against vessel walls. This collision between gas particles and vessel walls produces gas pressure.

Air is a mixture of gases: primarily nitrogen (N2; 78.6 percent), oxygen (O2; 20.9 percent), water vapor (H2O; 0.5 percent), and carbon dioxide (CO2; 0.04 percent). Each gas component of that mixture exerts a pressure. The pressure for an individual gas in the mixture is the partial pressure of that gas. Approximately 21 percent of atmospheric gas is oxygen. Carbon dioxide, however, is found in relatively small amounts (0.04 percent); therefore, the partial pressure for oxygen is much greater than that of carbon dioxide. The partial pressure of any gas can be calculated by: P = (Patm) (percent content in mixture).

Patm, the atmospheric pressure, is the sum of all of the partial pressures of the atmospheric gases added together: Patm = PN2 + PO2 + PH2O + PCO2= 760 mm Hg. The pressure of the atmosphere at sea level is 760 mm Hg. Therefore, the partial pressure of oxygen is: PO2 = (760 mm Hg) (0.21) = 160 mm Hg, while for carbon dioxide: PCO2 = (760 mm Hg) (0.0004) = 0.3 mm Hg. At high altitudes, Patmdecreases, but concentration does not change; the partial pressure decrease is due to the reduction in Patm .

Figure \(\PageIndex{1}\): Atmospheric pressure vs altitude: At high altitudes, there is a decrease in Patm, causing the partial pressures to decrease as well.

When the air mixture reaches the lung, it has been warmed and humidified within the nasal cavity upon inhalation. The pressure of the water vapor in the lung does not change the pressure of the air, but it must be included in the partial pressure equation. For this calculation, the water pressure (47 mm Hg) is subtracted from the atmospheric pressure: 760 mm Hg 47 mm Hg = 713 mm Hg, and the partial pressure of oxygen is: (760 mm Hg 47 mm Hg) 0.21 = 150 mm Hg.

These pressures determine the gas exchange, or the flow of gas, in the system. Oxygen and carbon dioxide will flow according to their pressure gradient from high to low. Therefore, understanding the partial pressure of each gas will aid in understanding how gases move in the respiratory system.