# Average number of degrees of freedom per molecule

1 vote
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When x amount of heat is given to a gas at constant pressure it perform x/3 amount of work. The average number of degrees of freedom per molecule is ?

I tried to do this by First law of thermodynamics as
$dQ=dU + dW$
$x-x/3 = dU$
but $dU = f/2 R dT$
so $2/3 x=f/2 R dT$

Now I am not able to find average number of degrees of freedom per molecule as temprature change is not given and also answer is not in terms of x but as integer.

Is the answer given as f=4

1 vote

From kinetic theory
$pV=NkT=\frac23(\frac32NkT)=\frac23K$
where $K$ is the total translational KE of the molecules of the gas, since each of the 3 degrees of translational motion has $\frac12 kT$ of energy on average.

The work done during expansion is
$p\Delta V=\frac23\Delta K=\frac13x$
$\Delta K = \frac12x$.

The amount by which the internal energy increases is $\Delta U=x-\frac13x=\frac23x$. This is shared equally between all $f$ degrees of freedom. The translational KE represents only 3 of these $f$ degrees of freedom. So by ratio
$\frac{3}{f}=\frac{\Delta K}{\Delta U}=\frac{\frac12x}{\frac23x}=\frac34$
$f=4$.

Alternatively :

The ratio of heat capacities is
$\frac{C_p}{C_v}=\gamma=1+\frac{2}{f}$.

Here $C_p \Delta T=x$ and $C_v \Delta T=\frac23x$ so
$\frac{C_p}{C_v}=\frac{C_p\Delta T}{C_v \Delta T}=\frac{x}{\frac23x}=\frac32=1+\frac24$.

Therefore $f=4$.

answered Apr 1, 2017 by (28,876 points)
edited Apr 1, 2017
@Sammy gerbil , I didn't get why KE represent 3 of these f degrees of freedom
why won't we count rotational or vibrational energy also?
I ought to have said that K is the **translational** KE of the molecules - ie $\frac12m<v^2>=\frac12m<v_x^2+v_y^2+v_z^2>=3.\frac12m<v_x^2>$. The other modes of vibration/rotation carry energy but it is only the translational KE which contributes to pressure, and pressure is how we judge temperature.

I have edited the text to make this point clearer.
$dQ = C_p dT=(f+2)RdT/2=x$
$(2/3)\frac{f+2}{2}RdT =f/2 RdT$