# Boyle's Law-$PV= nRT.$ What equation should be used to find pressure if n is not constant, like in an elastic system?

When air is added to an elastic system, like a balloon, the volume and pressure change. Would use Boyle's Law- $PV=nRT,$ but n does not remain constant in this situation.

Known: $V_1$ is at atmospheric pressure $(P_1)$. $V_2$ is known, but pressure is not $(P_2).$ What equation should be used to find the change in pressure?

• To be honest, the applicability of the ideal gas law cannot be easily justified in any environment. But, why are you assuming that T remains fixed? – Benjamin Mar 15 '16 at 4:57
• The equation PV=nRT would be valid at each instant of time even though n is changing (assuming that the change is slow enough that the system is at near-equilibrium at each instant of time). – user93237 Mar 15 '16 at 6:25
• @Benjamin I'm shocked by your answer. It seems so authoritative and categorical. All I can say is that you and I have very different views regarding the practical applicability of the ideal gas law. – Chet Miller Mar 15 '16 at 14:53
• @Chester. I am not saying that we should never use it. I am saying not always. Two features of this law is that (1) particles have zero size and (2) there is no interaction whatsoever between particles. This is approximately the case for very dilute gases whose molecules are not too big. But, we can assume these conditions hold and proceed with the solution. And agreeing with Samuel, you can use Boyle's Law only if you can justify (1) n is a very slow varying parameter and (2) you are going through isothermal paths. – Benjamin Mar 15 '16 at 16:47
• @Benjamin So, suppose you were faced with a practical situation in which you could potentially get your desired answer if you could use the ideal gas law, but you were not sure if it was "justified." How would you proceed? What would be your game plan? – Chet Miller Mar 15 '16 at 18:15

As stated the relationship cannot be solved. You have one equation, $PV = nRT$, and at least two unknowns.
Let's first assume the simple case where temperature is constant and the container you are introducing gas to is rigid and at some fixed volume $V_c$. Assume the gas in the container is initially at the same pressure as outside the container.
Now gas does have a springiness to it. In other words if you try to compress more molecules into a fixed volume, it stores the energy of compression and wants to 'push' back. If we further assume the ideal gas 'law' aka 'state equation', $PV=nRT$ is valid for our gas, then we can rearrange the equation as $$\frac{P}{n}=\frac{RT}{V}$$ If we let the volume in the denominator on the right side of the equation be the volume of the rigid container we can write $$\frac{P}{n}=\frac{RT}{V_c}$$ And the right side of the equation is a constant, and is known as the mass compliance. The compliance is not the compliance of the container, but rather the compliance of the volume within the container. What does this mean? Looking at the left side we see that this constant relationship of the container compliance is equal to pressure divided by volume. But we are interested in changes so we rather write $$\frac{\Delta P}{\Delta n}=\frac{RT}{V_c}$$ So if we force more molecules into the container, we raise the container pressure proportionally to the factor on the right side. But before we start forcing gas into the container we start with some molecules $n_o$ at the current atmospheric pressure (absolute) $P_o$.and so we can write $$\frac{\Delta P +P_o}{\Delta n + n_o}=\frac{RT}{V_c}$$ We further know that the mass is equal to the density of the gas times the volume of the gas $$n = \rho V$$ This can be substituted into the last equation getting $$\frac{\Delta P +P_o}{\rho (\Delta V + V_c)}=\frac{RT}{V_c}$$ finally we assume constant gas density, relative to the amount of compression getting $$\frac{\Delta P +P_o}{ \Delta V + V_c}=\rho \frac{RT}{V_c}=\frac{P_o}{V_c} = C$$