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Asked by manvirsingh2242 | 11 Jun, 2022, 09:14: AM
answered-by-expert Expert Answer
when two gases at same pressure and same temperature are allowed to mix ,
then final temperature will remain same as initial temperature.
 
hence this mixing process is isothermal expansion process for both the gases.
 
Change in entropy of isothermal expansion process, ΔS = n R ln( Vf / Vi )
 
where n is number of moles , R is universal gas constant , Vi is initial volume and Vf is final volume.
 
If N is number of molecules , n mole = N / NA  , where NA is avagadro number .
 
Change in entropy, ΔS =  N ( R / NA ) ln( Vf / Vi ) = N k ln( Vf / Vi )............................ (1)
 
where ( R / NA ) = k is Boltzman's constant
 
Vf = final volume = V1 + V2 , where V1 is initial volume occupied by N1 molecules of gas-1 and
 
V2 is initial volume occupied by N2 molecules of gas-2 .
 
Since at same temperature and pressure , volume is proportional to number of moles or number of molecules
 
we write eqn.(1) for entropy change of mixing of both the gases as
 
ΔS =   N1 k ln[ ( N1 + N2 ) / N2 ) + N2 k ln[ ( N1 + N2 ) / N2 )
 
if N1 ≈ N2 , we get , ΔS = ( N1 + N2 ) k ln(2)
Answered by Thiyagarajan K | 11 Jun, 2022, 16:43: PM
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