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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?

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

PLEASE DO CORRECT THIS ANSWER IF IT IS FAULTY.

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  • $\begingroup$ It looks good to me. $\endgroup$ – LDC3 Aug 31 '14 at 21:14

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