# Is ideal gas incompressible?

If so, I have read that for $$C_P$$ and $$C_V$$ for an incompressible gas are identical (in Heat Transfer by Yunus A. Cengel). Then how will the relation between $$C_P$$ and $$C_V$$ ($$C_P-C_V=R$$) hold for an ideal gas?

All gases, including ideal gases, are compressible. That's because the molecules of gases are far apart and can readily be brought together by pressure. The specific heats of gases depend on the process. For an ideal gas, $$C_p$$ and $$C_v$$ are the specific heats for constant pressure and volume processes, respectively, and are related by the equation $$C_{p}-C_{v}=R$$.

The molecules of liquids and solids are packed very closely together making it difficult to compress them. Only one specific heat is usually specified for a liquid or solid.

Hope this helps.

The state equation of an ideal gas $$pV=nRT$$ clearly shows it Is compressible.

Even more - infinitely compressible.

For an ideal gas, the heat required to raise the temperature at a constant pressure $$C_p$$ is more than work required at constant volume $$C_v$$, because in case of constant pressure, some work is also utilised in doing expansion work($$V$$ is not constant). In case of incompressible substance, there is no expansion work involved, so both specific heats are same.

Who said ideal gas is incompressible? Its (isothermal) compressibility is given by

$$\beta_{T}=-\dfrac{1}{V}\left(\dfrac{\partial V}{\partial P}\right)_{T}=-\dfrac{1}{V}\left(\dfrac{\partial}{\partial P}\dfrac{NK_{\rm B}T}{P}\right)_{T}=\dfrac{1}{V}\dfrac{NK_{\rm B}T}{P^{2}}=\dfrac{1}{P}$$

An ideal gas is compressible. I do not know what an incompressible gas is. Incompressibility implies that v is constant independent of pressure so trivially Cv=Cp.