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# A 9.380 mol sample of nitrogen gas is maintained in a 0.8198 L container at 301.8 K. What is the pressure in atm calculated using the van der Waals' equation for N2 gas under these conditions? For N2, a = 1.390 L2atm/mol2 and b = 3.910×10-2 L/mol. __ atm

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#### Explanation

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#### Explanation:

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Mar 2, 2018

$P = 331.35 \text{atm}$

#### Explanation:

The Van der Waals' equation states:

$\left(P + \frac{a {n}^{2}}{V} ^ 2\right) \left(V - n b\right) = n R T$

To solve for $P$, we can write:

$\left(P + \frac{a {n}^{2}}{V} ^ 2\right) = \frac{n R T}{V - n b}$

$P = \frac{n R T}{V - n b} - \frac{a {n}^{2}}{V} ^ 2$

Here,

$n = 9.38 \text{mol}$

$V = 0.8198 \text{L}$

$T = 301.8 \text{K}$

$a = 1.39 {\text{L"^2"atm mol}}^{-} 2$

$b = 3.91 \times {10}^{-} 2 {\text{L mol"^-1 "atm}}^{-} 1$

$R = 0.0821 {\text{L atm K"^-1"mol}}^{-} 1$

Inputting:

$P = \frac{9.38 \cdot 0.0821 \cdot 301.8}{0.8198 - 9.38 \cdot 3.91 \times {10}^{-} 2} - \frac{1.39 \cdot {9.38}^{2}}{0.8198} ^ 2$

$P = 331.35 \text{atm}$

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