Question

Derive the equation 1.40 from the equation 1.39

Short answer

The Equation 1.40 from the equation 1.39 is derived

as V^γ P = Constant.

Step-by-step solution

Step 1. Given data.

Equation 1.39 from the book as below

$V{T}^{f/2}=\text{constant.⋯⋯⋯⋯⋯⋯(1.39)}$.

Step 2. Derive Equation 1.40from1.39.

Relationship between Volume and temperature is given as

${\mathrm{VT}}^{\left(\frac{\mathrm{f}}{2}\right)}=Cons\tan t.................................\left(1\right)$

Expression for the ideal gas equation

$\begin{array}{rcl}\mathrm{PV}& =& \mathrm{NkT}\\ & & simplify\\ T& =& \frac{\mathrm{PV}}{\mathrm{Nk}}\end{array}$

Apply the power $\left(\frac{\mathrm{f}}{2}\right)$on both sides, we get

${\mathrm{T}}^{\left(\frac{\mathrm{f}}{2}\right)}={\left(\frac{\mathrm{PV}}{\mathrm{Nk}}\right)}^{\left(\frac{\mathrm{f}}{2}\right)}..............................\left(2\right)$

Multiplying equation$\left(2\right)$by V on both sides and simplify

$\begin{array}{rcl}{\mathrm{VT}}^{\left(\frac{\mathrm{f}}{2}\right)}& =& \mathrm{V}{\left(\frac{\mathrm{PV}}{\mathrm{Nk}}\right)}^{\left(\frac{\mathrm{f}}{2}\right)}\\ {\mathrm{VT}}^{\left(\frac{\mathrm{f}}{2}\right)}& =& {\mathrm{V}}^{1+\left(\frac{\mathrm{f}}{2}\right)}{\left(\frac{\mathrm{P}}{\mathrm{Nk}}\right)}^{\left(\frac{\mathrm{f}}{2}\right)}\end{array}$

Using equation$\left(1\right)$.

$$\begin{gathered} {\mathrm{VT}}^{\left(\frac{\mathrm{f}}{2}\right)}={\mathrm{V}}^{1+\left(\frac{\mathrm{f}}{2}\right)}{\left(\frac{\mathrm{P}}{\mathrm{Nk}}\right)}^{\left(\frac{\mathrm{f}}{2}\right)}=\text{constant} \\ {\mathrm{V}}^{1+\left(\frac{\mathrm{f}}{2}\right)}{\mathrm{P}}^{\left(\frac{\mathrm{f}}{2}\right)}=(\mathrm{Nk}{)}^{\left(\frac{\mathrm{f}}{2}\right)}·\text{constant.................................(3)} \end{gathered}$$

Here $\mathrm{N}.\mathrm{k},\mathrm{f}$are the constants.

$(\mathrm{Nk}{)}^{\left(\frac{\mathrm{f}}{2}\right)}·\text{constant}=\text{constant}$

${\mathrm{V}}^{1+\left(\frac{\mathrm{f}}{2}\right)}{\mathrm{P}}^{\left(\frac{\mathrm{f}}{2}\right)}=\text{constant}$

Step 3. Adiabatic exponent.

Apply the power $\left(\frac{2}{\mathrm{f}}\right)$

$\begin{array}{rcl}{\mathrm{V}}^{\left(\frac{2}{\mathrm{f}}\right)\left(1+\frac{\mathrm{f}}{2}\right)}{\mathrm{P}}^{\left(\frac{\mathrm{f}}{2}\right)·\left(\frac{2}{\mathrm{f}}\right)}& =& \text{constant}\\ {\mathrm{V}}^{\left(1+\frac{2}{\mathrm{f}}\right)}\mathrm{P}& =& \text{constant}\end{array}$

$1+\left(\frac{2}{\mathrm{f}}\right)=\left(\frac{\mathrm{f}+2}{\mathrm{f}}\right)$

Therefore,

${\mathrm{V}}^{\left(\frac{\mathrm{f}+2}{\mathrm{f}}\right)}\mathrm{P}=\text{constant}$

Defining the adiabatic exponent $\mathrm{\gamma }=\left(\frac{\mathrm{f}+2}{\mathrm{f}}\right)$the relation for a quasi-static adiabatic process

${\mathrm{V}}^{\mathrm{\gamma }}\mathrm{P}=\text{constant}$.

Hence the equation 1.40 is obtained from equation 1.39.