Chemistry IDEAL GAS EQUATION

Ideal Gas Law :

The results of the laws of Boyle and Gay-Lussac can be combined into an expression which represents the relationship between pressure, volume and temperature of a given mass of a gas; such an expression is described as an equation of state.

Suppose the gas is in the initial state with volume, `V_1`, pressure `p_1` and temperature `T_1`. We then change the state of the gas to a volume `V_2`, pressure `p_2` and temperature `T_2`. Let us carry out this change in two steps.

(i) first we change the pressure from `p_1` to `p_2` keeping the temperamre `T_1` constant.The resultant volume `V_1` as given by Boyle's law is

`V_r=(p_1V_1)/p_2`

(ii) Next, temperature is changed from `T_1` to `T_2`, keeping the pressure `p_2` constant. The final volume `V_2` as given by Charles law is

`V_2=(V_rT_2)/T_1=((p_1V_1//p_2)T_2)/T_1` or `(p_1V_1)/T_1=(p_2V_2)/T_2`

It follows that no matter how we change the state of the given amount of a gas, the ratio `pV//T` always remans constant, i.e.

`(pV)/T=K`

Value of `K` :

The value of `K` depends on the amount of the gas in the system. Since `V` is an extensive property (which is mass dependent), its value at constant `p` and `T` is proportional to the amount of the gas present in the system. Then `K` must also be proportional to the amount of gas because `p` and `T` are intensive properties (which have no mass dependence).

We can express this by writing `K = nR`, in which `n` is the amount of gas in a given volume of gas and `R` is independent of all variables and is, therefore, a universal constant. We thus have the general gas law

`pV=nRT`

The universal gas constant `R = (pV)/(nT)`. Thus, it has the unit of (pressure `xx` volume) divided by (amount of gas `xx` temperature). Now the dimensions of pressure and volume are,

Pressure = (force/area) = (force/ `text(length)^2` ) = force `xxtext(length)^(-2)`

Volume = `text(length)^(3)`

Thus `R= (text(force) xx text(length)^(-2) text(length)^3)/(text(amount of gas)(text(Kelvin)))`

`= ((text(force) xx text(length)))/((text(amount))(text(Kelvin)))`

`= (text[work(or energy)])/(text(amount of gas)(text(kelvin)))`

`text(Numerical Values of R)` :

i) In liter atmosphere = `0.0821` litre atm `text(deg)^(-1)` `text(mole)^(-1)`

ii) In ergs= `8.314xx 10^7` erg `text(deg)^(-1)` `text(mole)^(-1)`

iii) In calories = `1.987` cal `deg^(-1)` `text(mole)^(-1)`

iv) In joules = `8.314` J `text(deg)^(-1)` `text(mole)^(-1)`

`text(Use the value of R depending on the units in which value of pressure and volume has been used in ideal gas equation.)`

Relation between Molecular Mass and Gas Density :

From the ideal gas equation

`P = (n RT)/V = w/(M xx V) RT = (dRT)/M`

`M = d/P RT`

`M`= Molecular mass, `P` = Pressure, `T`=Temperature, `d`= Density

`text(Vapour Density)` : For gases another term which is often used is vapour density. Vapour density of a gas is defined as the ratio of the mass of the gas occupying a certain volume at a certain temperature and pressure to the mass of hydrogen occupying the same volume at the same temperature and pressure i.e.

`W_text(gas) = (PVM)/(RT)` and

`W_(H_2) = (PV xx 2)/(RT)` ( mol. wt. Hydrogen is `2`)

`W_(gas)/W_(H_2) =M/2 =` Vapour density of gas

VapourDensity `xx 2 =` Molecular wt.

`text(Vapour density of a gas is same at any temperature, pressure and volume.)`