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Given that $pV = 3600$, find the value of $\frac{dp}{dV}$ when $p=40$. I don't understand why the answer is $-\frac{4}{9}$. Can someone explain that?

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    or do you mean $$\frac{dV}{dp}$$? – 2017-01-28
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    No, I mean $\frac{dp}{dV}$ – 2017-01-28
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    Surely this can't be right, these variables are inversely proportional to each other so the derivative should be negative. – 2017-01-28
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    Oh sry, it should be -90 – 2017-01-28
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    I think ,it should be $\frac{-4}{9}$. – 2017-01-28
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    It should be $-40/90$ I suppose. – 2017-01-28
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    @SydneySniper Is now the given answer $-\frac{4}{9}$ ? Why you had posted $-90$ ? – 2017-01-28
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    Sorry there was an error I went and cross checked and asked my teacher. He said it should be $-\frac{4}{9}$ – 2017-01-28

5 Answers 5

4

$pV=3600$, so $V=\frac{3600}{p}$. Then $$\frac{dp}{dV}\cdot V + p\cdot \frac{dV}{dV}=0$$ $$\frac{dp}{dV}\cdot V = -p$$ $$\frac{dp}{dV} = -\frac{p}{V}=-\frac{40}{\frac{3600}{40}}=-\frac{4}{9}$$

3

For $p=40$ we have $V=\dfrac{3600}{40}=90$. Since $p=\dfrac{3600}{V}=3600V^{-1}$, it follows that $$ \dfrac{dp}{dV}=-\dfrac{3600}{V^2}=-\dfrac{3600}{90^2}=-\dfrac49. $$

2

By product rule

$$pV=3600\implies V\dfrac{dp}{dV}+p=0\implies\frac{dp}{dV}=-\frac pV$$

The answer should be $\displaystyle-\frac49$.

1

$p=\frac{3600}{V} \Rightarrow \frac{dp}{dV}= \frac{-3600}{V^2}$. Now when $p=40, V= 90$ so you should get $ \frac{dp}{dV}=\frac{-3600}{90^2}=\frac{-4}{9}$

1

Here $PV=3600$,then $P=\frac{3600}V$.Now $\frac{dP}{dV}=\frac{-3600}{V^2}$ but at $P=40,V=90.$So $\frac{dP}{dV}=\frac{-3600}{90^2}=\frac{-3600}{8100}=\frac{-4}{9}$