# Total and Partial Pressure - Dalton's Law of Partial Pressures

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**

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