Question

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a 9.327 mol sample of xenon gas is maintained in a 0.8435 l container at 299.3 k. what is the pressure in atm calculated using the van der waals equation for xe gas under these conditions? for xe, ( a = 4.194 \frac{l^{2} cdot atm}{mol^{2}} ) and ( b = 0.05105 \frac{l}{mol} ).
pressure = atm
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Explanation:

Step1: Recall van der Waals' equation

The van der Waals' equation is \((P + \frac{n^{2}a}{V^{2}})(V - nb)=nRT\). We need to solve for \(P\).
First, expand the left - hand side: \(P(V - nb)+\frac{n^{2}a}{V^{2}}(V - nb)=nRT\). Then, \(PV - Pnb+\frac{n^{2}a}{V}-\frac{n^{3}ab}{V^{2}}=nRT\). Solving for \(P\), we get \(P=\frac{nRT-\frac{n^{2}a}{V}+\frac{n^{3}ab}{V^{2}}}{V - nb}\)

Step2: Substitute the given values

Given \(n = 9.327\space mol\), \(V=0.8435\space L\), \(T = 299.3\space K\), \(a = 4.194\frac{L^{2}\cdot atm}{mol^{2}}\), \(b=0.05105\frac{L}{mol}\), and \(R = 0.0821\frac{L\cdot atm}{mol\cdot K}\)

• Calculate \(nRT\): \(nRT=9.327\times0.0821\times299.3\)

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• Calculate \(\frac{n^{2}a}{V}\): \(\frac{n^{2}a}{V}=\frac{9.327^{2}\times4.194}{0.8435}\)

$$
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• Calculate \(\frac{n^{3}ab}{V^{2}}\): \(\frac{n^{3}ab}{V^{2}}=\frac{9.327^{3}\times4.194\times0.05105}{0.8435^{2}}\)

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• Calculate \(nb\): \(nb=9.327\times0.05105 = 0.4762\)

• Calculate \(V-nb\): \(V - nb=0.8435- 0.4762=0.3673\)

• Now substitute into the \(P\) formula:

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Answer:

\(105.7\space atm\)