hiii tell me this questions right answer.
tell me fast at the spot don't tell the wrong answer....
What is the value of gamma for a
gas havin g 'n' degrees of freedom.
Answers
Answered by
3
ANSWER :
_________
Gamma, atomicity, γ = Cp / Cv
By derivation, Gamma, γ = 1 + 2 / f
Where, f = Degrees of freedom for 'n' number of molecules.
_________
Gamma, atomicity, γ = Cp / Cv
By derivation, Gamma, γ = 1 + 2 / f
Where, f = Degrees of freedom for 'n' number of molecules.
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Change in internal energy is independent of path followed? It is true. Why? sister please tell yar
Yes. It is true.
why
post it I'll describe
Bcz change in internal energy depends on it's final and initial stage not the path travelled.
Answered by
9
hi mate here is your answer....
Suppose, the number of degrees of freedom in a polyatomic gas molecule = n
Internal energy of one gram mole of the gas,
U = n * (1/2)kT * N
Where k = Boltzmann constant, T = temerature of system and N = Avogadro number
Or U = n/2 RT
Now we know,
Cv =dU/dT = > Cv = d/dT (U)
= (n / 2) R
Also we have a relation,
Cp = Cv + R
= ( n / 2) R + R
= (n/ 2 + 1) R
Again, Cp /Cv = Y
= ( n / 2 + 1) R / ( n / 2) R = ( 1 + 2 / n)
Therefore, Y = ( 1 + 2/ n)
Suppose, the number of degrees of freedom in a polyatomic gas molecule = n
Internal energy of one gram mole of the gas,
U = n * (1/2)kT * N
Where k = Boltzmann constant, T = temerature of system and N = Avogadro number
Or U = n/2 RT
Now we know,
Cv =dU/dT = > Cv = d/dT (U)
= (n / 2) R
Also we have a relation,
Cp = Cv + R
= ( n / 2) R + R
= (n/ 2 + 1) R
Again, Cp /Cv = Y
= ( n / 2 + 1) R / ( n / 2) R = ( 1 + 2 / n)
Therefore, Y = ( 1 + 2/ n)
thanks fir giving answer too much thanks
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