1
Lonely 1
·2009-04-10 09:06:08
Cv=3/2nRΔT
Cv=dU/dT=fR/2=R/γ-1
γ=Cp/Cv=1+2/f
f is the degree of freedom
21
tapanmast Vora
·2009-04-10 09:09:09
A more specific one dear....
Wich differentiates bw ACITVITY - NON-ACtivity OF VIBRATIONAL degrees of FREDDOM
1
Lonely 1
·2009-04-10 09:14:38
i think u r asking 4 the following
that the diatomic gas has 7 degrees of freedom(3tanslational,2rotational ,2vibrational)
f=6(all vibrations) at high temperatures
Actually the vibrational mode depends upon the temperature
4 JEE
u need to remember that
f=3 (translational mode)
f=5 (for diatomic)
or if a question comes they will ask degree of freedom or
it would be given
13
ДвҥїÑuÏ now in medical c
·2009-04-10 09:33:54
Now do this problem n1moles of monoatomic gas is mixed with n2moles of diatomic gas. find degrees of freedom of the mixture....
i like this one....
1
Lonely 1
·2009-04-10 09:43:39
n1+n2 = n1 + n2
---------- ----- ----
γ-1 γ1-1 γ1-1
for monoatomic gas f=3
so
f=1/γ-1
=>3=1/γ-1
γ1=4/3
γ2=6/5
put the values to get ur answer
24
eureka123
·2009-04-10 10:01:39
for mixture
CV=n1Cv1 +n2Cv2+...../n1+n2+....