Answer to Question #228851 in Molecular Physics | Thermodynamics for Unknown346307

Question

The following is the equation which connects u, p and v for several gases

u = a + bpv

where a and b are constants. Prove that for a reversible adiabatic process,

pv^γ = constant, where γ = b +1/b

Answer 1

For a reversible adiabatic process

$Q=0$

For a process of unit mass system

$dq=du+pdv$

or

$0=du+pdv$

We have

$u=a+bpv$

Differentiating both sides,

$du=bvdp+bpdv \\ 0=bvdp+bpdv+pdv \\ p(1+b)dv + bvdp=0$

Multiplying both sides by $\frac{1}{bvp}$ , we get

$\frac{1+b}{b} \times \frac{dv}{v} + \frac{dp}{p}=0$

Integrating both sides

$(\frac{1+b}{b})ln(v) +ln(p) = ln(C)$

Taking antilog

$pv^{(\frac{1+b}{b})}=C$

Using

$γ= \frac{1+b}{b}$

then

$pv^γ=C$

Proved.