Question

Prove that adiabatic proccess follows pv^(gamma)=c

Answer 1

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Answer 2

Answer:

From first law of thermodynamics :

dQ = dU + p dV

For adiabatic process dQ = 0

dU + p dV = 0

We can write dU / dt = C_v => dU = C_v dt

C_v dt + p dV = 0 .... ( i )

From ideal gas equation for one mole

p dV = RT

Now differentiate both side

p dV / dt + V dP / dt = R   [ Using product and power rule ]

dt = ( p dV + V dP ) / R

Putting dt value in ( i )

C_v ( p dV + V dP ) / R + p dV = 0

( C_v p dV +C_v V dP + R p dV ) / R = 0

C_v p dV + R p dV + C_v V dP  = 0

p dV ( C_v + R ) +  C_v V dP  = 0

We know :

C_p - C_v = R

C_p = C_v + R

p dV C_p +  C_v V dP  = 0

Now diving by C_v PV

p dV C_p /  C_v PV +  C_v V dP / C_v PV  = 0 /  C_v PV

C_p /  C_v ( dV / V ) + dP / P = 0

We know :

C_p /  C_v = γ

γ ( dV / V ) + dP / P = 0

Now integrating both side :

$\displaystyle{\gamma \int\limits {\dfrac{\text{dV}}{\text{V}} } + \int\limits {\dfrac{\text{dP}}{\text{P}} } = \text{C}}$

Using formula :

$\displaystyle{\int\limits {\dfrac{\text{dx}}{\text{x}}=\text{ln x }}}$

γ ln V + ln P = C

When ln is in addition we know it get multiplied and front one get into power :

$\displaystyle{\text{ ln PV}^{\gamma} = \text{C}}$

Now writing into simple exponent form :

$\displaystyle{\text{ PV}^{\gamma} = \text{e ^C }}$

$\display \text{Here e^C= K}$

$\displaystyle{\text{ PV}^{\gamma} = \text{K}}$

Therefore , proved.