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A van der Waal's gas obeys the equation ...

A van der Waal's gas obeys the equation of state `(P+(n^(2)a)/(V^(2)))(V-nb)=nRT`. Its internal energy is given by `U=CT-(n^(2)a)/(V^(2))`. The equation of a quasistatic adiabat for this gas is given by-

A

`T^(C//nR)V` constant

B

`T^((C//nR)//nR)V` = constant

C

`T^(C//nR)(V-nb)` = constant

D

`T^(C//nR)(V+nb)`= constant

Text Solution

Verified by Experts

The correct Answer is:
C
For adiabatic process dQ = 0 and `dU=dWrArrnC_vDeltaT=PDeltaV or nC_vdT=PdV` when change is very small
Now given `U=CT-(n^2a)/Vtherefore-DU=CdT+(n^2a)/V^2dV`
Put this value of dU in dU = dW
`therefore-(CdT+(n^2a)/V^2dV)=PdV" ...(i)"`
also `P=((nRT)/(V-nb))-(n^2a)/V^2` replace it in (i)
`-(CdT+(n^2a)/V^2dV)=(((nRT)/(V-nb))-(n^2a)/V^2)dV`
`thereforeCdT=((nRT)/(V-nb))dV`
`therefore-C/(nR)(dT)/T=(dV)/(V-nb)`
Integrating we get
In `T^(C//nR)=ln(V-nb)+k(krarr" constant of integration")`
`thereforeln(T^(C//nR))(V-nb)=k or (T^(C//nR))(V-nb)="constant"`
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