Consider the composite system, which is held at $300 K$ , shown in the following figure. Assuming ideal gas behavior, calculate the total pressure (in atm) if the barriers separating the compartments are removed. Assume that the volume of the barriers is negligible. (Given: $R = 0.082 a t m L / m o l . K)$

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Solution

The number of moles of $H e, N e$ and $X e$ is given by
$n_{H e} = \frac{P V}{R T} = \frac{3}{R T} n_{N e} = \frac{P V}{R T} = \frac{10}{R T}$
$n_{X e} = \frac{P V}{R T} = \frac{1}{R T}$
$n = n_{H e} + n_{N e} + n_{X e} = \frac{(3 + 10 + 1)}{R T} = \frac{14}{R T}$
Total volume of the vessel (V) = $2 + 4 + 1 = 7 L$
The total pressure is given by
$P = \frac{(n_{H e} + n_{N e} + n_{X e}) R T}{V}$
$P_{t o t a l} = \frac{(14)}{R T} \times \frac{R T}{7} = 2 a t m$

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Consider the composite system, which is held at 300 K, shown in the following figure. Assuming ideal gas behavior, calculate the total pressure (in atm) if the barriers separating the compartments are removed. Assume that the volume of the barriers is negligible. (Given: R=0.082 atm L/mol.K)

The number of moles of He, Ne and Xe is given by nHe=PVRT=3RTnNe=PVRT=10RT nXe=PVRT=1RT n=nHe+nNe+nXe=(3+10+1)RT=14RT Total volume of the vessel (V) = 2+4+1=7L The total pressure is given by P=(nHe+nNe+nXe)RTV Ptotal=(14)RT×RT7=2 atm