# Total and partial pressure - Dalton's law of partial pressures

## How to calculate total pressure and partial pressures for gas mixtures from Ideal Gas Law

The term **partial pressure** is used when we have a mixture of two or several gases in the same volume, and it expresses the pressure that is caused by each of the induvidual gases in the mixture.

The **total pressure** of the gas mixture is the sum of the partial pressure of the component gases:

P_{tot} = ∑P_{i} = P_{1} + P_{2} + P_{3} ...

Where

P_{tot} = the total pressure

P_{i} = the pressure of component* i* (*i* can vary from 1,2,3.....up to the number of different gases in the mixture)

From the Ideal Gas Law we have:

PV = nRT or P = nRT / V

Then,

P_{tot} = n_{tot}RT/V and P_{i} = n_{i}RT/V

where

n_{i} = the number of moles of component *i*

n_{tot} = the total number of moles in the gas mixture, which is the sum of all n_{i}.

R = the gas constant = 8.3145 [J/mol K] = 0.08206 [L atm/mol K] = 62.37 [L torr /mol K]

T = absolute temperature [K]

V = volume [m3] or [L]

For a gas mixture, the temperature and the volume is the same for all gases, and the gas constant is always the same, and then we get:

P_{i} /P_{tot} = (n_{i}RT/V)/(n_{tot}RT/V) = n_{i} /n_{tot}

We can express the concentration of one gas in the gas mixture as the mole fraction, X_{i}:

X_{i} = n_{i}/n_{tot}

and then

P_{i}/P_{tot} = X_{i} or P_{i} = X_{i}P_{tot}

See also Non-ideal gas - Van der Waal's equation and constants

### Example 1.

Dry air consists mainly of nitrogen (78.09vol% or 75.47wt%),oxygen (20.95vol% or 23.20 wt%), argon (0.93vol% or 1.28wt%) and carbondioxide (0.03vol% or 0.046wt%).

If you have 100 g of dry air in a 50 liter closed container, what will the partial pressure of each gas be, and what will the total pressure be at 120°C?

First, we must find how many moles of each gas, using the weight fraction of each gas and molweight of the gases:

n_{N2 }= 100[g] * 0.7547 /28.02 [g/mol] = 2.693 mol N_{2 }

n_{O2} = 100[g] * 0.2320 /32.00 [g/mol] = 0.725 mol O_{2 }

nAr = 100[g] * 0.0128 /39.95 [g/mol] = 0.032 mol Ar

n_{CO2 }= 100[g] * 0.00046 /44.01 [g/mol] = 0.001 mol CO_{2}

n_{tot} = n_{N2} + n_{O2} + n_{Ar} + n_{CO2} = *3.451 mol gas*

Then, assuming the gas mixture behaves ideally, we have:

**The total pressure, P _{tot}** = n

_{tot}RT/V = 3.451 [mol]* 0.08206 [L atm/mol K]* (273+120) [K] / 50 [L] =

**2.226 atm**

**P _{N2}**= X

_{N2}*P

_{tot}= n

_{N2}/n

_{tot}*P

_{tot}= (2.693[mol]/3.451[mol])*2.226 atm =

**1.737 atm**

**P _{O2}**= X

_{O2}*P

_{tot}= n

_{O2}/n

_{tot}*P

_{tot}= (0.725[mol]/3.451[mol])*2.226 atm =

**0.468 atm**

**P _{Ar}**= X

_{Ar}*P

_{tot}= n

_{Ar}/n

_{tot}*P

_{tot}= (0.032[mol]/3.451[mol])*2.226 atm =

**0.021 atm**

**P _{CO2}**= X

_{CO2}*P

_{tot}= n

_{CO2}/n

_{tot}*P

_{tot}=(0.001[mol]/3.451[mol])*2.226 atm =

**0.0006 atm**