States Of Matter

CHARLES LAW

It states "at constant pressure, the volume of a given mass of a gas, increases or decreases by \( \frac{1} {{273.15}}\) th of its volume at 0ºC for every rise or fall of one degree in temperature".
       \( \frac{{V_t }} {{V_0 }}\) = 1 + \( \frac{1} {{273.15}}\)t     (at constant n and P)
        or     V_t = V₀  \( \left( {1 + \frac{t} {{273.15}}} \right)\)
        or     V_t = \( \frac{{V_0 (273.15 + t)}} {{273.15}}\)
 0ºC on the Celsius scale is equal to 273.15 K at the Kelvin or absolute scale. 
  i.e. T_t (Temperature in Kelvin scale) = 273.15 + t
    \(\therefore\) From the above equation we get  \( \frac{{V_t }} {{V_0 }} = \frac{{T_t }} {{T_0 }}\) 
        or    \( \frac{{V_t }} {{T_t }} = \frac{{V_0 }} {{T_0 }}\) 
 i.e.  The volume of a given gas is proportional to the absolute temperature.
                    \(\frac {V_1} {T_1}=\frac {V_2} {T_2}\)  (at constant P)

States Of Matter

*  Graphical Representation

   
                       
Graphs between V and T at constant pressure are called Isobars.

Ex. If the temp. of a particular amount of gas is increased from 27ºC to 57ºC, find final volume of the gas, if initial volume = 1 lt and assume pressure is constant.

States Of Matter

SOLVED EXAMPLES

Ex.  If the temp. of a particular amount of gas is increased from 27ºC to 57ºC, find final volume of the gas, if initial volume = 1 lt and assume pressure is constant.
Sol.        
         \(\frac{1} {{(273 + 27)}} = \frac{{V_2 }} {{(273 + 57)}} \)   So      V₂ = 1.1 lt. 

Ex. An open container  of volume 3 litre contains air at 1 atmospheric pressure. The container is heated from initial  temperature 27ºC or 300 K to tºC or (t + 273) K the amount of the gas expelled from the container measured 1.45 litre at 17ºC and 1 atm.Find temperature t.
Sol.      \(\therefore\)  T⁰₁ = 300 K
It can be assumed that the gas in the container was first heated to (t + 273), at which a volume ‘\(\Delta\)V’ escaped from the container 
        hence applying charles law : 
            \(\frac{3} {{300}} \)  = \(\frac{{3 + \Delta V}} {{t + 273}}\)
Now, this volume ‘DV’ which escapes when the container get cooled 
    \(\therefore\)    \(\frac{{\Delta V}} {{t + 273}}\) = \(\frac{{1.45}} {{290}}\)
Solve the two equations and get the value of \(\Delta\)V and t.
determine  \(\Delta\)V & calculate t that will be the answer. 

States Of Matter

Gay-Lussac's Law :
Dependence of Pressure on Temperature
It states "at constant volume, the pressure of a given mass of a gas is directly proportional to the absolute temperature of the gas".
          P \( \propto \) T    or    P = KT
        or    \( \frac{{P_1 }} {{T_1 }} = \frac{{P_2 }} {{T_2 }}\)

* Graphical Representation 
        
                        
The plots drawn at constant volume for a gas is called as Isochore.

States Of Matter

The Combined Gas Law :
It states "for a fixed mass of gas, the volume is directly proportional to absolute temperature and inversely proportional to the pressure".
Boyle's Law, V \( \propto \frac{1} {P}\) (at constant n, T)
Charle's Law, V  \( \propto \) T (at constant n, P)
Therefore, V \( \propto \frac{T} {P}\)    or     V = K\( \frac{T} {P}\)       or     \( \frac{{PV}} {T} = K\)     or   \( \frac{{P_1 V_1 }} {{T_1 }} = \frac{{P_2 V_2 }} {{T_2 }}\)   

Volume Coefficient (\(\alpha\)_v) of a Gas :
 The ratio of increase in volume of a gas at constant pressure per degree rise of temperature to its volume at 0ºC is the volume coefficient of the gas.
    \(\alpha\)_v = \( \frac{{V_t -V_0 }} {{V_0 \times t}}\)     or      V_t = V₀ (1 + \(\alpha\)_vt)
For all gases, \(\alpha\)_v = \( \frac{1} {{273}}\)

Pressure Coefficient (\(\alpha\)_p) of a Gas :
The ratio of increase in pressure of the gas at constant volume per degree rise of temperature to its pressure at 0ºC is the pressure coefficient of the gas.
    \(\alpha\)_p = \( \frac{P_ t -P_0 } {{P_0 \times t}}\)         or     P_t = P₀ (1 + \(\alpha\)_pt)
For all gases, \(\alpha\)_p = \( \frac{1} {{273}}\)