# Deriving simple equation for volume and pressure, adiabatic exponent

I have

$$VT^\frac{f}{2}=\text{constant}\tag1$$

and

$$V^{\gamma}P=\text{constant}.\tag2$$

Where $$V=\text{volume}, \ P=\text{pressure} , \ T=\text{temperature},$$ $$f=\text{number of degrees of freedom per molecule}$$ and $$\gamma=\frac{f+2}{f}.$$

I want to derive $$(2)$$ from $$(1)$$

Using the ideal gaslaw I have $$T=\frac{PV}{Nk}$$, where $$N$$ and $$k$$ are constant. Plugging this into $$(1)$$ I get

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

and a constant times a constant is constant so finally

$$VT^{\frac{f}{2}}=V^{\frac{f+2}{2}}P=\text{constant}.$$

However my exponent is off, the denominator in the exponent should be $$f$$ but I get $$2$$. What am I missing?

• Please state reason for downvote. – Parseval Sep 11 '19 at 14:52
• Seems to be a very 'silly' question; if not question, atleast your response to my answer is not genuine – Tojrah Sep 13 '19 at 12:26

You did not put $$P^{f/2}$$ after the first step. From there you will get the right answer (then you will take both sides of the equation to the 2/f power).
$$V\bigg(\frac {PV}{Nk} \bigg)^{f/2}=\frac{P^{f/2} V^{1+f/2}}{constant}$$
• $P$ is not supposed to have any exponent in the final answer. – Parseval Sep 11 '19 at 13:59