Adiabatic Process: Proving Variation of Gamma

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An adiabatic process is characterized by no heat transfer, represented by the equation PV^γ = constant, where γ is the heat capacity ratio. For diatomic gases, γ is approximately 1.4, while for monoatomic gases, it is about 1.6. The value of γ is defined as the ratio of specific heat at constant pressure (Cp) to specific heat at constant volume (Cv). The theoretical proof of these values can be derived from the principles of heat capacity and the behavior of ideal gases during reversible adiabatic processes. Understanding these concepts clarifies the significance of γ in thermodynamics.
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I know that adibatic process means no heat transfer.

i.e., →PV\gamma =constant.

Where \gamma = 1.4 for diatomic gas, \gamma= 1.6 for monoatomic gas.

My question is how \gamma=Cp/Cv ?

And can we prove theoritcally that \gamma = 1.4 for diatomic gas, \gamma= 1.6 for monoatomic gas.
 
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ajayguhan said:
I know that adibatic process means no heat transfer.

i.e., →PV\gamma =constant.

Where \gamma = 1.4 for diatomic gas, \gamma= 1.6 for monoatomic gas.

My question is how \gamma=Cp/Cv ?

Gamma is defined as the ratio of the specific heat at constant pressure to the specific heat at constant temperature.

And can we prove theoritcally that \gamma = 1.4 for diatomic gas, \gamma= 1.6 for monoatomic gas.

http://en.wikipedia.org/wiki/Heat_capacity_ratio
 
ajayguhan said:
I know that adibatic process means no heat transfer.

i.e., →PV\gamma =constant.

This equation is only true for an ideal gas undergoing a reversible (i.e., no entropy generation) adiabatic process.

ajayguhan said:
My question is how \gamma=Cp/Cv ?

It is simply a definition for which we found a physical significance (Like work is defined as a force times its displacement).

ajayguhan said:
And can we prove theoritcally that \gamma = 1.4 for diatomic gas, \gamma= 1.6 for monoatomic gas.

Read The simple case of the monatomic gas and Diatomic gas on Theory of heat capacity.
 
Thank you for spending time to clarify my doubts. Your answer helped me.
 
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