Question:
Root mean square velocity of molecules of a diatomic gas is increased to 1.5 times. If the process is done adiabatically, the ratio of initial volume to the final volume is.
a)2.25
b)4.6
c)5.6
d)7.6
Also, in the above question if the process is done isobarically, then the required ratio is
a)16/27 b)8/9 c)4/9 d)2/5
And the ratio of work done by the gas during adiabatic and isobaric process is. (reference to above questions)
a)1
b) -1
c) 2.5
d) -2.5
Answer:
There are 3 parts in this problem (a) Finding the initial volume by final volume for adiabatic process
(b) Finding the initial volume by final volume for isobaric process
(c) Finding the ratio of work done by adiabatic to work done by isobaric process
Root mean square velocity of a diatomic gas = √3RT/M
For diatomic gas CP = 7/2 and Cv = 5/2; Hence Υ = CP / Cv = 7/5
Let the initial and final volume be Vi and Vf respectively; Similarly, the initial and final temperature be T1 and T2 respectively.
RMS velocity initial = √3RTi/M ---------Eqn (1)
RMS velocity final = √3RTf/M ----------Eqn (2)
Dividing 2 by 1 and the problem first line states it equal to 1.5
Hence, ratio √Tf /Ti = 1.5 ; (Tf /Ti) = (1.5)2 ---------Eqn (3)
Case 1:
By adiabatic equation, we know TiViΥ-1 = TfVfΥ-1 ------------------Eqn (4)
Hence, Tf/Ti = (Vi/Vf)Υ-1
Υ = CP / Cv = 7/5; Hence Υ -1= 2/5
Tf/Ti = (Vi/Vf)Υ-1
Using Eqn (3). we get (1.5)2 = (Vi/Vf)2/5
(Vi/Vf) = (1.5)5 = 7.59
Case 2:
By isobaric process, the pressure remains constant. So in the ideal gas eqn PV = RT or V is proportional to T ------------------Eqn (5)
Initial volume / Final volume = (Vi/Vf) = (Ti/Tf)
By Eqn (3), (Vi/Vf) = (Ti/Tf) = 1/(1.5)2 = 1/(3/2)2 = 4/9
(Vi/Vf) = 4/9
Case 3:
By adiabatic process, Work done = R(T2 – T1) / 1-Υ
By isobaric process, Work done = nR(T2 – T1) / 1-Υ
Hence, work one by adiabatic / work done by isobaric = 1/ 1-Υ = 1/(2/5) = 5/2 = 2.5
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