Problem in understanding the Proof of $PV^{\gamma}$ =constant in thermodynamics 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)
 A: 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.
A: We know that for an isochoric process $C_v= (\frac{\partial U}{\partial T})_v$; i.e. $dU=C_vdT$ at constant volume.
Now first law of thermodynamic is $\bar{d}Q=dU+dW;$ For adiabatic process $\bar{d}Q=0$.
$\therefore$  $dU=-dW$  or, $dU=-pdV$
Now $dU=C_vdT;$                                                                                  
If we allow to change the volume of the system then the system doing a work against the surrounding pressure and for this work we have to supply heat form out side. 
In case of adiabatic process the energy for work provides the internal energy of the gas and that's why the temperature of the gas reduces. Now this reduce of internal energy must be, $dU=C_vdT;$
So, $dU=C_vdT=-pdV$
If we rise the temperature of the gas by $dT$ keeping volume constant then we have to supply $C_vdT$ amount of heat which would increase the internal energy of the gas by $dU$. Now if I expand this gas adiabatically and the temparature of the gas reduced by $dT$ then we can say that the externally supplied heat in isochoric process is converted in work to expand the gas. So the change of internal energy $dU$ must be $C_vdT$  
A: The proof for the adiabatic process can be understood much more simply if you use calculus, specifically differentials of multivariable functions, as demonstrated by the answer from Rajesh Sardar. 
For example, $W=-P\Delta V$ is incorrect in this case since pressure isn't constant during the process (a fact you can check by simply looking at an adiabat in a P-V diagram). It's just a special case (for the isobaric process) of the general formula $\delta W=-pdV$ and the first law of thermodynamics. The standard theory of thermodynamics also explains why for an ideal gas we always have $U=C_VT$, even when the process isn't isochoric.
What you are trying to do (derive the formula for the adiabatic process without calculus) is very hard, if not impossible. I suggest you delve more deeply into the mathematics of it all and try figuring out the problem in a few years.
