Answer to Question #220347 in Molecular Physics | Thermodynamics for Happiness Edobo

Question

Show that p1v1^gamma =p2v2^gamma

Answer 1

Answer:-

We know from the 1st law of thermodynamics

$dq=du+dw$

For an adiabatic process, we can say dQ=0,

so, $dU+dW=0=dU+PdV......1$

We know, $dU=nCvdT.....2$

For an ideal gas, 

$PV=nRT$

Or, $T={PV\over nR}$

or, $dT={PdV+VdP\over nR}$

Putting in 2,we get,

$dU={C_v\over R}(PdV+VdP)$

Putting this value of dU in 1 we get,

${dP\over P}=−{dV\over V}{C_v+R\over C_v}$

or, ${dP\over P}=−{dV\over V}⋅γ$ (as $γ={Cp\over Cv}={Cv+R\over Cv}$ )

or,$∫{dP\over P}=−γ∫{dV\over V}$

or, $lnP=−γlnV+c$

or, $lnP+γlnV=k$

or, $lnP+lnV^γ=k$

or,$ln(PV^γ)=k$

so, $PV^γ = constant$

So it can also be written as

$P_1V_1^γ=P_2V_2^γ$