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For an ideal gas at constant temperature, the variables $p$ (pressure) and $v$ (volume) are related by the equation $pv^k=c$, where $k$ and $c$ are constants. If the volume is changed slightly from $v$ to $v+Δv$, what quadratic approximation expressing $p$ in terms of $Δv$ would you use?

(Find the approximation valid for $Δv≈0$.)

problems with letters are definitely not my fad.
could anyone please help?

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  • $\begingroup$ $p=c/v^k$, and use Taylor expansion $\endgroup$ – Michael Galuza Jul 17 '15 at 12:44
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When $v=v_0+\Delta v$,

$$p = c v^{-k} = c v_0^{-k} \left (1+\frac{\Delta v}{v_o} \right )^{-k} = c v_0^{-k} \left (1-k \frac{\Delta v}{v_o} + \frac{(-k) (-k -1)}{2!} \left (\frac{\Delta v}{v_o} \right )^2 + O \left (\frac{\Delta v}{v_o} \right )^3 \right )$$

So, as a quadratic approximation, we have

$$p \approx c v_0^{-k} - c k v_0^{-k-1} \Delta v + \frac12 c k (k+1) v_0^{-k-2} \left (\Delta v\right )^2 $$

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Hint: Think of $p$ as a function of $v$: $$p(v) = \frac{c}{v^k}.$$ The quadratic approximation around $v$ would be: $$p(v+\Delta v) \approx p(v) + p'(v)\Delta v + \frac{p''(v)}{2} (\Delta v)^2$$

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Possibly something like trying to expand $p(v+\Delta v)^{k}=c$ with \begin{eqnarray} (v+\Delta v)^{k} &=& \sum_{j=0}^{k} {k \choose j}v^{k-j}\Delta v^{j} \\ &=& v^{2}+2v\Delta v+\Delta v^{2}+\ldots \end{eqnarray} Since $\Delta v \approx 0$ we could simply write this as $v^{2}+2v \Delta v$.

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