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College Chemistry : Non-Ideal Gas Behavior

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Example Question #1 : Non Ideal Gas Behavior

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

Possible Answers:

$\displaystyle 0.732 atm$

$\displaystyle 1.37 atm$

$\displaystyle 1.73 atm$

$\displaystyle 0.841 atm$

$\displaystyle 1.00 atm$

Correct answer:

$\displaystyle 1.37 atm$

Explanation:

The pressure of the flask at $100^{\circ}C$ $\displaystyle 100^{\circ}C$ is $\displaystyle 1.37 atm$.

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

Because the volume between the flasks and the moles in each flask are constant, we can cancel out $\displaystyle V$ and $\displaystyle n$.

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

At STP, conditions are $\displaystyle 1.00 atm$ and $\displaystyle 273.15 K$.

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

$P_{2}= 1.37 atm$ $\displaystyle P_{2}= 1.37 atm$

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Example Question #3 : Gases

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.

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

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

Possible Answers:

$3.45\textrm{ atm}$ $\displaystyle 3.45\textrm{ atm}$

$7.43\textrm{ atm}$ $\displaystyle 7.43\textrm{ atm}$

$7.47\textrm{ atm}$ $\displaystyle 7.47\textrm{ atm}$

$6.42\textrm{ atm}$ $\displaystyle 6.42\textrm{ atm}$

$6.90\textrm{ atm}$ $\displaystyle 6.90\textrm{ atm}$

Correct answer:

$6.90\textrm{ atm}$ $\displaystyle 6.90\textrm{ atm}$

Explanation:

Recall the van der Waals equation for non-ideal gases

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

Rewrite the equation using the known values and solve for P

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

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

$\textrm{P = 6.90 atm}$ $\displaystyle \textrm{P = 6.90 atm}$

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College Chemistry : Non-Ideal Gas Behavior

Study concepts, example questions & explanations for College Chemistry

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Example Questions

Example Question #1 : Non Ideal Gas Behavior

Example Question #3 : Gases

All College Chemistry Resources

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