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For an adiabatic process, the ideal gas follows the equation

$$ PV^{\gamma}= constant$$

The equation above implies that the pressure of an ideal gas (under adiabatic process) depends on the "degree of freedom" of that gas.

But if pressure depends on degree of freedom then

why isn't this factor used in the ideal gas equation ($PV=nRT$) itself ?

Or

Why does the pressure exerted by a gas depend on it's degree of freedom in an adiabatic process ?

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  • $\begingroup$ I think I get what you are asking. The thing is an ideal gas, if it is in equilibrium, it must obey $PV=nRT$. The $PV^{\gamma}=C$ is an additional independent constraint on top of that for the state variable which comes when the gas under goes an adiabatic process. $\endgroup$ Jul 21, 2022 at 13:26
  • $\begingroup$ You should see the derivation the $\gamma$ comes because in an adiabatic process, the work infinitesimal equals the internal energy infinitesimal. Now, we know that the ideal gas has also a constraint that the internal energy is dependent on degree of freedom always, so in this special case it becomes that the $PdV$ is dependent on the coefficient itself $\endgroup$ Jul 21, 2022 at 13:28

2 Answers 2

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In the kinetic model of an ideal gas, the pressure exerted on the walls is due to the molecules' collisions with the walls. This means that pressure is solely a function of the energy stored in the molecules' translational degrees of freedom; internal degrees of freedom do not affect it.

On the other hand, in an adiabatic process, any work done on the gas must go into the thermal energy of the gas. What's more, the equipartition theorem says that the thermal energy of the gas must be shared between each type of degree of freedom equally. Taken all together, this implies that for a given amount of work done, less energy will go into translational degrees of freedom when the molecules have more internal degrees of freedom. This explains why the pressure depends on the number of degrees of freedom when we're talking about an adiabatic process.

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It depends on the number of degrees of freedom of a molecule of the ideal gas, which enters in the expression for the energy of the ideal gas, from which the adiabatic equation is derived: $$ U=\frac{D}{2}nk_BT,\\ dU=dQ - PdV $$ Here $D$ is the number of the degrees of freedom.

For an adiabatic process $dQ=0$, and we obtain $$ \frac{D}{2}nk_BdT=PdV=\frac{nk_BT}{V}dV $$ Integratinga nd converting back to $PV$ variables one obtains the equation for the adiabatic process.

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