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The ideal gas law states that for a given quantity of confined gas in a container, the pressure P, volume V and temp T satisfy an equation $P=kT/V$, where k is a positive constant. Show that:

$$V \frac{\partial P}{\partial V} = -P$$

and

$$V \frac{\partial P}{\partial V}+T\frac{\partial P}{\partial T} = 0.$$

I would appreciate anyones explanation to the two parts of this question, after many tries the closest I can come to is =P for the first part and cannot find my mistake.

Ian
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E.JJ
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  • Can you write your attempt so we can point out your mistake? (Also, I will put MathJax into your question; try to imitate it when you expand on your attempt.) – Ian Oct 04 '15 at 18:40
  • from P=kT/V that simplifies to V=kT/P, then taking the partial derivative of P with repsect to V of (kT/V) = kT/V^2 and subbing that with the value of V, yields kT/(kT/P) = P – E.JJ Oct 04 '15 at 18:45
  • The derivative of $1/x$ is $-1/x^2$. – Ian Oct 04 '15 at 19:01

1 Answers1

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More generally, let $y=cx^\alpha$. Then

$$ x\frac{\partial y}{\partial x}=xc\alpha x^{\alpha-1}=\alpha cx^\alpha=\alpha y\;. $$

In your case, we have $P\sim V^{-1}$ and $P\sim T^1$, so

$$ V\frac{\partial P}{\partial V}+T\frac{\partial P}{\partial T}=-1\cdot P+1\cdot P=0\;. $$

joriki
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