Non-Ideal Gas Behavior

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AP Chemistry › Non-Ideal Gas Behavior

Questions 1 - 2

1

One flask is at STP and another is at $100^{\circ}C$ . What is the pressure at $100^{\circ}C$ ?

Explanation

The pressure of the flask at $100^{\circ}C$ is .

$\frac{P_{1}V_{1}}{n_{1}T_{1}}=\frac{P_{2}V_{2}}{n_{2}T_{2}}$

Because the volume between the flasks and the moles in each flask are constant, we can cancel out and .

$\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$

At STP, conditions are and .

$\frac{1.00 atm}{273.15K}=\frac{P_{2}}{373.15K}$

$P_{2}= 1.37 atm$

2

A 3.00 L container at 273 K is filled with 1.00 mol Cl2(g), which behaves non-ideally.

Using the van der Waals equation, calculate pressure exerted by the gas.

$a=6.49\frac{\textrm{L}^{2}\textrm{atm}}{\textrm{mol}^{2}}$

$b=0.0562\frac{\textrm{L}}{\textrm{mol}}$

$6.90\textrm{ atm}$

$6.42\textrm{ atm}$

$7.47\textrm{ atm}$

$7.43\textrm{ atm}$

$3.45\textrm{ atm}$

Explanation

Recall the van der Waals equation for non-ideal gases

$\textrm{(P + }\frac{n^{2}a}{\textrm{V}})(\textrm{V - }nb)=nR\textrm{T}$

Rewrite the equation using the known values and solve for P

$[\textrm{P + }\frac{(1\textrm{ mol})^{2}(6.49\frac{\textrm{L}^{2}\textrm{ atm}}{\textrm{mol}^{2}})}{(3.00\textrm{ L})^2}][\textrm{3.00 L - (1 mol)}(0.0562\frac{\textrm{L}}{\textrm{mol}})]=(1\textrm{ mol})^{2}(0.0821\frac{\textrm{atm}}{\textrm{mol K}})(273K)$

$(\textrm{P + 0.721 atm})(2.94\textrm{ L})=22.4\textrm{ atm L}$

$\textrm{P = 6.90 atm}$

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