Adiabatic Processes

Now we take a look at processes that happen so fast that no heat exchange with the environment can take place. The change in the inner energy is then given as $$\mathrm{d} U = -p\,\mathrm{d}V$$ Replacing $\mathrm{d}U$ with $c_\mathrm{V}\,\mathrm{d} T$ and inserting the ideal gas equation leads to $$c_\mathrm{V}\frac{\mathrm{d} T}{T} = -T\frac{\mathrm{d} V}{V}$$ Integrating both sides gives $$c_\mathrm{V} \ln T = R\ln V + \mathrm{const}$$ Rearranging this equation leads to $$\ln\left(T^{c_\mathrm{V}}V^{-R}\right) = \mathrm{const}$$ Using the definition of $c_p$ results in $$T^{c_\mathrm{V}} V^{c_\mathrm{V} - c_\mathrm{p}} = \mathrm{const}$$ Solving this equation leads to $$\boxed{TV^{\mathrm{\kappa-1}} = \mathrm{const}}$$ Here, the adiabatic index $\kappa$ has been defined as $$\boxed{\kappa = \frac{c_\mathrm{p}}{c_\mathrm{V}}}$$ Together with the ideal gas equation, the following correlation with the pressure and volume can be written as: $$\boxed{p V^\kappa = \mathrm{const}}$$ For an ideal gas containing single particles, the degree of freedom is $f=3$ and hence $\kappa=(f+2)/f = 5/2$. For more complicated objects such as molecular oxygen or nitrogen, it will be $7/2$. In general, the adiabatic curve decreases much faster than an isothermal curve for the same conditions.

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