# Addition of combined gas law constant

There are two containers connected to each other but separated by a cork. If container(1) has a gas $G_1$ at a volume $V_1$, pressure $P_1$ and temperature $T_1$ and similarly in container(2) there is a gas $G_2$ at volume $V_2$, pressure $P_2$ and temperature $T_2$. If the cork between the containers is removed and the temperature changed to $T_3$ what is the pressure of the mixture inside the system of containers.

My textbook gives it as:

# $\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}=\frac{P_3V_3}{T_3}$

where $V_3$ = $V_1$+$V_2$

# $P_3=\frac{T_3}{V_3}\left(\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}\right)$

I don't understand how that comes out. Can we just add the constants(of combined gas law) like that?

• There are no constants in the equations you posted. – LDC3 Aug 31 '14 at 21:08

Caution: This answer is exclusively from my perspective.

Moments after uploading this question I realized the answer myself. For those who strike upon this question. I 'think' the answer goes like this:

The combined gas law states that :

$\frac{PV}{T}=nR$ ---------- [or $PV=nRT$ ideal gas equation] (where 'n' is the number of moles of the gas in consideration)

Now therefore:

$\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}$ = $n_1R+n_2R$ = $R\left(n_1+n_2\right)$

Now

$n_1+n_2=n_3$ (NOTE: $n_3$ is not the number of moles of a single gas but the gases $G_1$ and $G_2$ combined)

This can be said as the volume gets added up. Therefore further referring to Avogadro's Law (of equal volume) we the latter equation yields the equation in question.