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$R$ is the universal gas constant, the value of which you can find e.g. here. For this problem, you don't need to know the value, though. $$ p V = n R T \mathrm{\ \ \ or\ \ \ \ } R = \frac{p V}{n T}$$ is true for any ideal gas, and a pretty good approximation for most gases at room temperature or higher and atmospheric pressure or lower. So for two different ...


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Yes, of course. Rearrange the ideal gas law into the form $$\frac{P}{T} = \frac{nR}{V}.$$ Since by construction the volume and number of moles is fixed, the right-hand side of the equation is constant, so the pressure increases linearly with the temperature in your situation.


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For an ideal gas, $pV/(RT) = n$ can explain a lot of situations because it contains four variables. If you add gas to an empty Mylar balloon, the volume will increase. If it is still not filled to maximal volume, heating the gas will increase the volume. If you have a full balloon and gently step on it, the pressure will increase; same if you add more gas to ...


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Only for ideal gases the equation $pV = nRT$ holds good. Therefore $pV/RT = n$ is valid only for ideal gases. And in the graph, $n$ is constant for ideal gas. For the real gases $\ce{CH4}$, $\ce{N2}$, $\ce{H2}$, $\ce{CO2}$, while the amount of substance $n$ is kept constant, the ratio $pV/RT$ is not a constant anymore ($pV/RT \ne n$). Instead, for real ...


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The whole point of the graph is that $pV/(RT)$ is no longer equal to $n$ for a real gas. $n$ isn't changing (as you point out, this is physically impossible), but the ratio $pV/(RT)$ is changing to some value that is not the same as $n$. Once you're freed from the ideal gas equation, there's no need for $n$ to "follow" the changes in pressure / ...


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