A gas undergoes a process such that `pprop1/T`. If the molar heat capacity for this process is `C=33.24J//mol-K`, find the degree of freedom of the molecules of the gas.

Answer

4

Explanation

As `pprop1/T`
or `pT=` constant …(i)
We have for one mole of an ideal gas
`pV=RT` …(ii)
From Eqs. (i) and (ii),
`p^2V=` constant
or `pV^(1//2)=K (say) =p_iV_i^(1//2)=p_fV_f^(1//2)` …(iii)
From first law of thermodynamics,
`DeltaQ=DeltaU+DeltaW`
or `CDeltaT=C_VDeltaT+DeltaW`
or `C=C_V+(DeltaW)/(DeltaT)` ...(iv)
Here, `DeltaW=intpdV=Kint_(V_i)^(V_f)V^(-1//2)dV`
`=2K[V_f^(1//2)-V_i^(1//2)]=2[p_fV_f^(1//2)V_f^(1//2)-p_iV_i^(1//2)V_i^(1//2)]`
`=2[p_fV_f-p_iV_i]=2R[T_f-T_i]`
`=(RDeltaT)/(1//2)implies(DeltaW)/(DeltaT)=2R`
Substituting in Eq. (iv), we have
`C=C_V+2R=(R)/(gamma-1)+2R`
Substituting the value, `33.24=R((1)/(gamma-1)+2)=8.31((1)/(gamma-1)+2)`
Solving this we get `gamma=1.5`
Now, `gamma=1+2/F`
or degree of freedom `F=(2)/(gamma-1)=(2)/(1.5-1)=4`