• BME 200

# Dalton’s Law

Dalton’s Law says that the pressure of a mixture of ideal gases is the sum of the partial pressures of each gas in the mixture. The partial pressure of a gas is the pressure that gas would exert on a container if it were by itself. Air is made up of a mixture of gases and each gas contributes its own partial pressure.

$$P_\mathrm{air}=P_\mathrm{O_2}+P_\mathrm{CO_2}+P_\mathrm{N_2}+P_\mathrm{H_2O}+P_\mathrm{other}=760~\mathrm{mmHg}$$

where $$P_\mathrm{O_2}$$ is the partial pressure of oxygen, $$P_\mathrm{CO_2}$$ is the partial pressure of carbon dioxide, and so on. Since $$P_\mathrm{other}$$ is a very small fraction of $$P_\mathrm{air}$$, it is often neglected from partial pressure calculations.

In dry air (i.e., no water vapor), the values are

$$P_\mathrm{air}=160+0.3+600+0=760~\mathrm{mmHg}$$

So, in dry air, the partial pressure of $$O_2$$ ($$P_{O2}$$ is 160 mmHg. It’s also useful to reprsent the partial pressure as a fraction. In this case, the partial pressure of oxygen would be $$\frac{160}{760}$$ = 0.21.

Inhaled (alveolar) air has water vapor in it, and since the total pressure must remain at 760 mmHg, the other partial pressures must change slightly. In other words

$$P_\mathrm{alv}=100+40+559+47 = 760~\mathrm{mmHg}$$

In alveolar air, the partial pressure of oxygen is much lower, carbon dioxide is higher, nitrogen is lower, and water vapor is higher, compared to dry atmostpheric air.

# Relation to mass balance

Since we can use the partial pressure to know how much oxygen (or carbon dioxide) is being inhaled or exhaled, we can use the ideal gas low to figure out what mass of oxygen and carbon dioxide are being exchanged during ventilation. The ideal gas law is

$$PV=nRT$$

where $$P$$ is the gas pressure, $$V$$ is the gas volume, $$n$$ is the number of moles, $$R$$ is the universal gas constant, and $$T$$ is the temperature in Kelvin.

## Example

Assume the alveolar volume of a single breath is 350 mL. What is the exhaled mass of oxygen?

We assume STP, so rearranging the ideal gas law we have

$$n=PV/RT=\frac{\frac{160}{760}(101325~\textrm{Pa})~(0.35~\textrm{cm}^3)}{(8.314\times 10^6~\textrm{cm}^3~\textrm{Pa}~^\textrm{o}\textrm{K}^{-1}~\textrm{mol}^{-1})(273~^\textrm{o}\textrm{K})} = 3.288 \times 10^{-6}~\textrm{mol}$$

Since the molar mass of atmospheric oxygen is 32 g/mol, the mass of oxygen exhaled is

$$m_{O2}=(3.288 \times 10^{-6}~\textrm{mol})(32~\textrm{g}~\textrm{mol}^{-1}) = 0.1~\textrm{mg}$$

Last updated:
August 25, 2018