I have looked at the proof of this relation $PV^\gamma = C$; (where $P$ is pressure and $V$ is volume) in quite some places but I am not able to understand the logic behind the third step.
In REVERSIBLE ADIABATIC expansion-
$W= -P\Delta V$
If $\Delta T$ is the fall in temperature then
$C_V\Delta T = -P\Delta V$, where $C_V$ means specific heat at constant volume. This is where I am having problem. I know here adiabatic expansion is there so $q=0$ that is $\Delta U$ or $\Delta E = W$ ,but how can we write $W = C_V\Delta T$? (Doubt 1)
Also even though I don't understand how this is written but I still know $C_V$ is heat capacity at constant volume but volume is changing here(expansion), so how can we use $C_V$ which is meant to be used at a specific volume only?? (Doubt 2)
Best Answer
The internal energy of a mono-atomic gas is given by:
$E_{\text{int}}=\dfrac{3}{2}nRT$.
Where $n$ is the number of moles and $R$ is the gas constant and $T$ is the temperature.
The statement of conservation of energy is given by:
$E_{\text{int}}=Q+W$
The work done by the gas is given by $W=-\int PdV$.
For a gas undergoing temperature variation at constant volume, the work $W$ done by it, is of course zero, therefore
$\Delta E_{\text{int}}=Q=\dfrac{3}{2}nR\Delta T$.
Now we define the molar heat capacity at constant volume $C_V$ as the amount of heat $Q$ required to raise the temperature of one mole of a gas by $1$ degree at constant volume. Therefore it follows from the last equation that:
$C_V=\dfrac{3}{2}R$ for one mole of any monoatomic gas.
For an adiabatic process applied to one mole of a gas, by definition $Q=0$ therefore it follows from $E_{\text{int}}=Q+W$ that
$\Delta E_{\text{int}}=W=\dfrac{3}{2}R\Delta T=-P\Delta V$.
Since $\dfrac{3}{2}R=C_V$.
Therefore
$C_V\Delta T = -P\Delta V$.
For any given mono-atomic gas, since it's always the case that $C_V=\dfrac{3}{2}R$, therefore you can use it whenever you like, whether the process under question is isovolumetric or not.