# Why gamma is not equal to 2?

In my text book, I came across a ratio called heat capacity ratio or gamma which is equal to

$$\gamma = 1+\frac{2}{f}$$

where $$f$$ is degree of freedom. Therefore

$$C_p = \gamma C_v$$

where $$C_p$$ and $$C_v$$ are molar heat capacity at constant pressure and constant volume.

In an online lecture I came to know that energy given to a piston gas system is equally shared by piston and gas (when piston is movable).Amount of energy gained by piston is work done and by gas is change in internal energy.let us assume that change in internal energy dU cause change in temperature by 1 kelvin. let us say that we have 1 mole of gas in a cylinder piston system So Cv is amount of energy needed to change temperature of our system by 1 kelvin when piston is immovabe ( constant volume) therefore Cv=dU and Cp be energy needed to do the same temperature change when piston is movable (constant pressure)so Cp=2dU in later scenario Cp is equally shared by gas and piston therefore Cp must equal to 2Cv since Cv =dU .By definition I came to know that R is amount of energy needed to rise temperature of one mole of substance by one kelvin so Cv must equal to R(gas constant) since Cv also defined in the same way .But in my book It is said that gamma is greater than one and less than two (usually).why degree of freedom affect Cp Cv relation and Cv R relation? why Cp is not equal to 2 Cv ? why Cv is not equal to R ?why with increase in temperature gamma value decrease?

• I recommend that you edit your question for readability. I've made a partial edit to get you started. – Alfred Centauri Sep 3 '19 at 11:22
• Thanks for the edit I don't have computer to edit like this .I am asking this questions in an old smart phone. – Kamal Sep 3 '19 at 11:46
• Can you provide a source for that claim? An idealized piston should not absorb any energy from the system during a quasistatic process like this; the additional energy is transferred to the environment as work, and the amount of work done is generically not simply half of the heat added to the system. – J. Murray Sep 3 '19 at 11:49

Your mistake is you attribute 1 degree of freedom to the gas. In an ideal gas there are three degrees of freedom for every particle in the gas because each particle can move in three dimensions. The piston does have an equal share of the energy, but it only gets one degree of freedom for its ability to move in one dimension. Since there are about 1 mole of gas atoms, one more or less degree of freedom is completely irrelevant.

The $$f$$ in the equation $$\gamma = 1 + \frac{2}{f}$$ is degrees of freedom per particle in the gas.

Other answers have done a good job of dealing with the details of the equations, so I'm not going to reproduce them.

• An ideal gas has a minimum of 3 degrees of freedom for a monatomic (noble) gas. There are 5 for diatomic and 7 for a polyatomic gases. – Bob D Sep 3 '19 at 13:06
• @BobD My understanding was that non-monotomic gases were considered non-ideal since the number of degrees of freedom there is actually temperature dependent (see, for example, the graphs in this question) – Sean E. Lake Sep 3 '19 at 15:03
• That is not true. Most gases can be considered ideal provided the temperature is not too low and the pressure is not too high. For example, air at STP can be treated as an ideal gas within reasonable tolerances, and air consists primarily of two diatomic gases. I don't see anything in the link you provided that says diatomic and polyatomic gases can't exhibit ideal gas behavior. – Bob D Sep 3 '19 at 15:39
• @BobD It depends on how strictly you define "ideal". If you just define it as "obeys $PV=NkT$", then you're right. If you include "Number of degrees of freedom are temperature independent," then not as much. Admittedly, still a very close approximation since the heat capacities vary slowly with temperature for every non-noble gas. – Sean E. Lake Sep 3 '19 at 15:45
• Sean, yes I meant in terms of obeying the ideal gas law. I think that's what one normally means by ideal gas behavior. As far as "number of degrees of freedom are temperature interdependent" sure, since the specific heats are temperature dependent it follows that gamma is temperature dependent, and it follows from that the $f$ in the equation for gamma must be temperature dependent. My only reason for my initial comment was it seemed your answer implied ideal gases only have 3 degrees of freedom, which is not the case. – Bob D Sep 3 '19 at 15:53

In piston/cylinder gas problems the piston is usually considered massless and simply acts as the mechanism for the gas to do work on the surroundings, and there is no "energy gained" by the piston. In any case, if you are dealing with an ideal gas you have made several incorrect assumptions.

So Cv is amount of energy needed to change temperature of our system by 1 kelvin when piston is immovable ( constant volume) therefore Cv=dU

For an ideal gas, $$\Delta U=C_{v}\Delta T$$ for any process not just for a constant volume (immovable piston) process.

Cp be energy needed to do the same temperature change when piston is movable (constant pressure)so Cp=2dU

For an ideal gas, this is not correct. As indicated above, $$\Delta U=C_{v}\Delta T$$ for any process.

By definition I came to know that R is amount of energy needed to rise temperature of one mole of substance by one kelvin so Cv must equal to R(gas constant)

This is also not correct. For an ideal gas $$C_{v}=C_{p}-R$$. So if you wanted to express the change in internal energy in terms of $$C_{p}$$, it would be

$$\Delta U=(C_{p}-R)\Delta T$$

Since your initial assumptions are incorrect, any conclusions you reach based upon them will be incorrect.

Finally, the minimum number of degrees of freedom for a gas is 3 (monatomic gas), in the x-y-z directions. That means the maximum value of gamma is 1.667.

Hope this helps.

• In my question dq=dU+dw and dq=Cp and dw=du therefore Cp = du+du=2du – Kamal Sep 3 '19 at 12:52
• @Kamal First, for a constant pressure process $dq=C_{p}dT$. Secondly, where does $dw=dq$ come from? What happened to $dU$? And if $du=0$ then $dq=dw$. – Bob D Sep 3 '19 at 12:57
• I am assuming that magnitude of change in internal energy is same as magnitude of workdone – Kamal Sep 3 '19 at 13:21
• @Kamal If that's the case then the process has to be adiabatic which means $dq=0$ and not $dq=C_p$ – Bob D Sep 3 '19 at 13:40