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I recognise my question is at a beginner level but my current level of knowledge of math is up to what any undergraduate engineer would know, so you can give me a more-than-beginner-level explanation.

I'm puzzled by why exactly our rules for algebraic manipulation are deemed valid in equations like the ideal gas law or Ohm's law (to give a few examples) where everything contained in the equation are variables. Why is it valid to proceed with the mentality of "having the same on both sides" here?

For instance if I have $pV=nRT$ which is the ideal gas law, I can say $p=\frac{nRT}{V}$ following the well-known rules for manipulating equations.

I think this type of manipulation differs from solving an equation like $5x=15$ to get $x=3$ in that in that case you are looking for a value that "exists" or "is some how contained inside x" and in the case of an equation like the ideal gas law, the variables inside (except R) can be any positive real number, so a variable is kind of another set of entity which contains all numbers inside the letter; which makes it difficult to grasp how you can, i.e., "divide by V on both sides" if V is not anything at all.

Thanks, and I really hope someone can help out of this confusion because it's really hampering my ability to study at the moment.

ben ari
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  • For one thing, physically, $pV=nRT$ and $p=\frac{nRT}{V}$ are just different ways of saying the same thing. You are using math to change the way you look at it. In the "real world" these variables are all real, fixed values. The variables and equation manipulation/math only comes about when we are studying it, or figuring out what we haven't already measured. My $0.02 – turkeyhundt Jan 14 '15 at 01:22

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I take it that if you had $2p=nRT$, you'd be comfortable dividing each side by 2, and if you had $3p=nRT$, you'd be comfortable dividing by 3, etc.

The goal is to say all those things at once. That is, we want to say "If $2p=nRT$, then $p=nRT/2$. Also, if $3p=nRT$, then $p=nRT/3$. Also..... ".

Another way to say this "No matter what the value of $V$ might be, if $pV=nRT$, then $p=nRT/V$". That's exactly what the algebra allows you to say. And saying it this way is far more efficient, because the other way you're never done saying it.

WillO
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