Question

derive the equation of adiabatic change​

Answer 1

AnSwer :-

i) Consider 1g mole of an ideal gas enclosed in cylinder fitted with a perfectively friction less position. Let P, v, T be initial pressure, volume and temperature.

Suppose a small amount ofheat de spent in two ways :

Increasing the temperature of gas by a small range dT at constant Volume = C$\sf _v$dT

ii) Exansion of gas by a small volume d portion of heat spent = PdV

dθ = C$\sf _v$dT + Pdv

In adiabatic change,

dθ = 0

C$\sf _v$dT + PdV = 0

According to standard gas equation

PV = RT

differentiate both sides

PdV + VdP = RdT

$\sf dT = \dfrac{PdV + VdP}{R} \: \: \: .....(1)$

putting in e.q (1)

$\sf \dfrac{C_p (PdV + VdP)}{R} + PdV = 0$

$\sf (C_v + R)PdV + C_vVdP = 0$

$\sf C_p - C_v = R \: \: \: .....(2)$

from (2) dividing both sides by C$\sf _p$PV

$\sf \dfrac{C_pPdV}{C_pPV} + \dfrac{C_pPdV}{C_pPV} = 0$

$\sf \dfrac{dV}{V} + \dfrac{dP}{P} = 0$

$\sf \gamma = C_p /C_v$

Integrating both sides

$\sf \int \dfrac{dV}{V} + \int \dfrac{dP}{P} = C$

$\sf \gamma log_{e}V + log_{e}P = C$

$\sf log_{e} {V}^{ \gamma } + log_{e}P = C$

$\sf log_{k}P {V}^{ \gamma } = C$

$\sf PV^{ \gamma } = C$

$\:$

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