Chemistry Ideal Gas Equation and Dalton's Law of Partial Pressures

Topics Covered :

● Ideal Gas Equation
● Density and Molar Mass of a Gaseous Substance
● Dalton's Law of Partial Pressures
● Partial Pressure in Terms of Mole Fraction

Ideal Gas Equation :

`text(Definition :)` The three laws which we have learnt till now can be combined together in a single equation which is known as ideal gas equation.

● At constant `T` and `n; V prop 1/p` Boyle’s Law

● At constant `p` and `n; V ∝ T` Charles’ Law

● At constant `p` and `T ; V ∝ n` Avogadro Law

● Thus `V prop (nT)/p` ...........(15)

`=> V = R (nT)/p` ...........(16)

where `R` is proportionality constant.

● On rearranging the equation (16) we obtain

`pV = n RT` .............(17)

`=> R = (p V)/(n T)` ..................(18)

● `R` is called gas constant.

● It is same for all gases.

● Therefore, it is also called `text(Universal Gas Constant)`.

● Equation (17) is called `text(ideal gas equation)`.

`=>` Equation (18) shows that the value of `R` depends upon units in which `p`, `V` and `T` are measured.

`=>` If three variables in this equation are known, fourth can be calculated.

● From this equation we can see that at constant temperature and pressure, `n` moles of any gas will have the same volume because

`V = ( n RT)/p`

and `n`, `R`, `T` and `p` are constant.

● This equation will be applicable to any gas, under those conditions when behaviour of the gas approaches ideal behaviour.

● Volume of one mole of an ideal gas under STP conditions (`273.15 K` and `1` bar pressure) is `22.710981 L mol^(-1)`.

● Value of `R` for one mole of an ideal gas can be calculated under these conditions as follows :

`R = ((10^5 pa) (22.71 ×10^(–3)m^3))/((1 mol) (273.15 K))`

`= 8.314 Pa m^3 K^(–1) mol^(–1)`

`= 8.314 × 10^(–2)` bar `L K^(–1) mol^(–1)`

`= 8.314 J K^(–1) mol^(–1)`

● At STP conditions used earlier (`0 °C` and `1` atm pressure), value of `R` is `8.20578 × 10^(–2) L atm K^(–1) mol^(–1).`

● Ideal gas equation is a relation between four variables and it describes the state of any gas, therefore, it is also called `text(equation of state)`.

● Ideal gas equation is the relationship for the simultaneous variation of the variables.

`=>` If temperature, volume and pressure of a fixed amount of gas vary from `T_1`, `V_1` and `p_1` to `T_2`, `V_2` and `p_2` then we can write

`(p_1 V_1)/T_1 = nR` and `(p_2 V_2)/T_2 = nR`

`=> (p_1V_1)/T_1 = (p_2 V_2)/T_2` .........(19)

● Equation (19) is a very useful equation.

● If out of six, values of five variables are known, the value of unknown variable can be calculated from the equation (19).

● This equation is also known as `text(Combined gas law)`.
Q 3056223174

At `25°C` and `760 mm` of Hg pressure a gas occupies `600 mL` volume. What will be its pressure at a height where temperature is `10°C` and volume of the gas is `640 mL`.


`p_1 = 760 mm Hg , V_1 = 600 mL`

`T_1 = 25+273 = 298 K`

`V_2 = 640 mL` and `T_2 = 10+273 = 283K`

According to Combined gas law

`(p_1 V_1)/T_1 = (p_2 V_2)/T_2`

`=> p_2 = ( p_1 V_1 T_2)/(T_1 V_2)`

`=> p_2 = (760 mm xx600 mL xx 283 K)/(298K xx640 mL)`

` = 676.6 mm Hg`

Density and Molar Mass of a Gaseous Substance :

Ideal gas equation can be rearranged as follows :

`n/V = p/(RT)`

Replacing `n` by `m/M` we get

`m/(MV) = p/(RT)` ................(20)

`d/M = p/(RT)` (where d is the density) ......................... (21)

On rearranging equation (21), we get the relationship for calculating molar mass of a gas.

`M = (d RT)/p` ...........(22)

Dalton’s Law of Partial Pressures :

`=>` The law was formulated by John Dalton in 1801.

`text(Statement :)` The total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases i.e., the pressures which these gases would exert if they were enclosed separately in the same volume and under the same conditions of temperature.

● In a mixture of gases, the pressure exerted by the individual gas is called `text(partial pressure)`.

● Mathematically,

`p_text(Total) = p_1+p_2+p_3.....` ( at constant `T`, `V`) ............(23)

where `p_text(Total)` is the total pressure exerted by the mixture of gases and `p_1`, `p_2`, `p_3` etc. are partial pressures of gases.

`=>` Gases are generally collected over water and therefore are moist.

● Pressure of dry gas can be calculated by subtracting vapour pressure of water from the total pressure of the moist gas which contains water vapours also.

● Pressure exerted by saturated water vapour is called `text(aqueous tension)`.

● Aqueous tension of water at different temperatures is given in Table.

`p_text(dry gas) = p_text(Total) -` Aqueous tension ...........(24)

Partial pressure in terms of mole fraction :

● Suppose at the temperature `T`, three gases, enclosed in the volume `V`, exert partial pressure `p_1`, `p_2` and `p_3` respectively.


`p_1 = (n_1 RT)/V` .............(25)

`p_2 = (n_2 RT)/V` .........(26)

`p_3 = (n_3 RT)/V` .............(27)

where `n_1 n_2` and `n_3` are number of moles of these gases.

● Thus, expression for total pressure will be

`p_text(Total) = p_1 + p_2 + p_3`

`= n_1 (RT)/V + n_2 (RT)/V + n_3 (RT)/V`

`= ( n_1 + n_2 + n_3) (RT)/V` .........(28)

● On dividing `p_1` by `p_text(total)` we get

`p_1/p_text(total) = (n_1/(n_1+n_2+n_3))`

` = n_1/(n_1+n_2+n_3)`

where `n = n_1 + n_2 + n_3`

`x_1` is called mole fraction of first gas.

Thus `p_1 = x_1 p_text(total)`

Similarly, for other two gases we can write

`p_2 = x_2 p_text(total)` and `p_3 = x_3 p_text(total)`

Thus, a general equation can be written as `p_1 = x_1 p_text(total)` ........(29)

where `p_i` and `x_i` are partial pressure and mole fraction of `ith` gas respectively. If total pressure of a mixture of gases is known, the equation (29) can be used to find out pressure exerted by individual gases.
Q 3016323279

A neon-dioxygen mixture contains `70.6 g` dioxygen and `167.5 g` neon. If pressure of the mixture of gases in the cylinder is `25` bar. What is the partial pressure of dioxygen and neon in the mixture ?


Number of moles of dioxygen

` = (70.6 g)/{32 g mol^(-1)}`

`= 2.21 mol`

Number of moles of neon

` = (167.5 g)/{20 g mol^(-1)}`
` = 8.375 mol`

Mole fraction of dioxygen

` = (2.21)/(2.21+8.375)`

`= (2.21)/(10.585)`

` = 0.21`

Mole fraction of neon ` = (8.375)/(2.21+8.375)`

`= 0.79`

Partial pressure of a gas = mole fraction × total pressure

⇒ Partial pressure of oxygen = 0.21 × (25 bar) = = 5.25 bar

Partial pressure of neon = 0.79 × (25 bar) = 19.75 bar