Gaseous State

Root mean square (rms) velocity of N₂ molecule (u) = $\sqrt {{{3RT} \over M}} $.

For N₂ molecule, molecular mass, M = 28. Therefore,

$u = \sqrt {{{3RT} \over {28}}} $ ..... (1)

When the temperature becomes doubled, that is, 2T. Let the new root mean square (rms) velocity of N-atom be u'. For N-atom, the molecular mass, M = 14. Therefore,

$u' = \sqrt {{{3RT} \over {14}}} $....... (2)

On dividing Eq. (1) by Eq. (2), we get

${u \over {u'}} = \sqrt {{1 \over 4}} $

$ = {1 \over 2} \Rightarrow u' = 2u$

According to van der Waals' equation, for one mole of a gas

$\left( {p + {a \over {{V^2}}}} \right)(V - b) = RT$ ...... (1)

At high pressure, ${a \over {{V^2}}}$ can be neglected

So, $p + {a \over {{V^2}}} \approx p$ ...... (2)

From Eqs. (1) and (2), we get

$p(V - b) = RT$

$pV - pb = RT$

$pV = RT + pb$

Dividing both the sides by RT, we get

${{pV} \over {RT}} = 1 + {{pb} \over {RT}}$

$Z = 1 + {{pb} \over {RT}}$

For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant. $i.e$

$V \propto n$

Hence in the given case.

Initial moles and final moles are equal ${\left( {{n_T}} \right)_i} = {\left( {{n_T}} \right)_f}$

${{{P_i}V} \over {R{T_1}}} + {{{P_i}V} \over {R{T_1}}}$

$ = {{{P_f}V} \over {R{T_1}}} + {{{P_f}V} \over {R{T_2}}}$

$2{{{P_i}} \over {{T_1}}} = {{{P_f}} \over {{T_1}}} + {{{P_f}} \over {{T_2}}}$

$ \Rightarrow {P_f} = {{2\,{P_i}{T_2}} \over {{T_1} + {T_2}}}$

Compressibility factor $\left( Z \right) = {{PV} \over {RT}}$

(For one mole of real gas)

van der Waals equation

$\left( {P + {a \over {{V^2}}}} \right)\left( {V - b} \right) = RT$

At low pressure, volume is very large and hence correction term $b$ can be neglected in comparison to very large volume of $V.$

i.e.$\,\,\,V - b \approx V$

$\left( {P + {a \over {{V^2}}}} \right)V = RT;$

$PV + {a \over V} = RT$

$PV = RT - {a \over V};$

${{PV} \over {RT}} = 1 - {a \over {VRT}}$

Hence, $\,\,\,z = 1 - {a \over {VRT}}$

Most probable speed $\left( {{C^ * }} \right) = \sqrt {{{2RT} \over M}} $

Average Speed $\left( {\overline C } \right) = \sqrt {{{8RT} \over {\pi M}}} $

Root mean square velocity $\left( c \right) = \sqrt {{{3RT} \over M}} $

${C^ * }:\overline C :C = \sqrt {{{2RT} \over M}} :\sqrt {{{8RT} \over {\pi M}}} :\sqrt {{{3RT} \over M}} $

$ = 1:\sqrt {{4 \over \pi }} :\sqrt {{3 \over 2}} $

$ = 1:1.128:1.225$

$\left( {P + {a \over {{V^2}}}} \right)\left( {V - b} \right) = RT\,\,$

at high pressure ${a \over {{V^2}}}$ can be neglected

$PV - Pb = RT\,\,\,$

and $\,\,\,PV = RT + Pb$

${{PV} \over {RT}} = 1 + {{Pb} \over {RT}}$

$z = 1 + {{Pb} \over {RT}};Z > 1\,\,\,$ at high pressure

The value of $a$ is a measure of the magnitude of the attractive forces between the molecules of the gas. Greater the value of $'a',$ larger is the attractive inter-molecular force between the gas molecules.

The value of $b$ related to the effective size of the gas molecules. It is also termed as excluded volume.

The gases with higher value of $a$ and lower value of $b$ are more liquefiable, hence for $C{{\rm l}_2}$ $''a'''$ should be greater than for ${C_2}{H_6}$ but for it $b$ should be less than for ${C_2}{H_6}.$

From the ideal gas equation :

$PV = nRT\,\,$

or $\,\,\,n = {{PV} \over {RT}} = {{3170 \times {{10}^{ - 3}}} \over {8.314 \times 300}}$

$ = 1.27 \times {10^{ - 3}}$

Let the mass of methane and oxygen $=m$ $gm.$

Mole fraction of ${O_2}$

$ = {{Moles\,\,of\,\,{O_2}} \over {Moles\,\,of\,\,{O_2}\, + \,Moles\,\,of\,\,C{H_4}}}$

$ = {{m/32} \over {m/32 + m/16}}$

$ = {{m/32} \over {3m/32}} = {1 \over 3}$

Partial pressure of ${O_2}$

$=$ Total pressure $ \times $ mole fraction of ${O_2},$

${P_{{O_2}}} = P \times {1 \over 3} = {1 \over 3}P$

Distribution of molecular velocities at two different temperature is given shown below.



NOTE : At higher temperature more molecules have higher velocities and less molecules have lower velocities.

As evident from fig. thus it is clear that With the increase in temperature the most probable velocity increase but the fraction of such molecules decreases.

In van der Waals equation $'b'$ is for volume correction.

${{K.E\,\,of\,\,neon\,\,at\,\,{{40}^ \circ }C} \over {K.E\,\,of\,\,neon\,\,at\,\,{{20}^ \circ }C}}$

$ = {{{3 \over 2}K \times 313} \over {{3 \over 2}K \times 293}}$

$ = {{313} \over {293}}$

According to kinetic theory the gas molecules are in a state of constant rapid motion in all possible directions colloiding in a random manner with one another and with the walls of the container and between two successive collisions molecules travel in a straight line path but show haphazard motion due to collisions.

Value of gas constant

$\left( R \right) = 0.0821L\,atm\,{K^{ - 1}}\,mo{l^{ - 1}}$

$ = 8.314 \times {10^7}\,\,ergs\,{K^{ - 1}}\,mo{l^{ - 1}}$

$ = 8.314J{K^{ - 1}}\,mo{l^{ - 1}}$

$ = 1.987\,cal\,{K^{ - 1}}\,mo{l^{ - 1}}$

$PV = nRT\,\,$ (number of moles $ = n/V$)

$\therefore$ $\,\,\,n/V = P/RT.$

Kinetic theory of gases proves all the given gas laws.