Question

A gas undergoes adiabatic transformation using the first law of thermodynamics show that this process is represented by the equation pVy =constant

Answer 1

The first law of thermodynamics says

$$\delta Q=dU+pdV$$

For adiabatic processes

$$\delta Q=0$$

Hence

$$dU+pdV=0$$

$$\left(\frac{\partial U}{\partial T}\right)_VdT+\frac{\left[\left(\frac{\partial U}{\partial V}\right)_TdT+p\right]}{\left(\frac{\partial T}{\partial V}\right)_p}dV=0$$ $$C_VdT+\frac{\left[C_p-C_V\right]}{\left(\frac{\partial T}{\partial V}\right)_p}dV=0$$

For perfect gas

$$pV=RT$$

$$C_VdT+\frac{\left[C_p-C_V\right]}{\frac{V}{T}}dV=0$$

$$\frac{dT}{T}+\left[C_p/C_V-1\right]\frac{dV}{V}=0$$

$$\frac{dT}{T}+\left[\gamma-1\right]\frac{dV}{V}=0$$

After integrating

$$\ln T+(\gamma-1)\ln V=\ln \rm Const$$

$$TV^{\gamma-1}=\rm Const$$

or

$$\frac{pV}{R}V^{\gamma-1}=\rm Const$$

Finally

$$pV^{\gamma}=\rm Const$$