Partial Pressure - College Chemistry

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Question

A mixture of helium, nitrogen, and neon has a total pressure of $815\text{ mmHg}$ at a temperature of $255\text{K}$ . If the partial pressure of helium is $201\text{ mmHg}$ and the partial pressure of nitrogen is $351\text{ mmHg}$ , what mass of neon is present in the mixture?

Answer

Since we have the total pressure and the pressures of helium and nitrogen, we can find the pressure of neon in this container.

$\text{P}_{tot}=\text{P}_{He}+\text{P}_{N}_2+\text{P}_{Ne}$

$\text{P}_{tot}-\text{P}_{He}-\text{P}_{N}_2=\text{P}_{Ne}$

$\text{P}_{Ne}=815-201-351=263\text{ mmHg}$

Next, convert this pressure into atmospheres.

$263\text{ mmHg}(\frac{1\text{ atm}}{760\text{mmHg}})=0.34605\text{ atm}$

Now, recall the Ideal Gas Law:

Rearrange the equation to solve for moles of gas:

$n=\frac{PV}{RT}$

Using the pressure of neon, find the moles of neon that is present in the container.

$n=\frac{(0.34605\text{ atm})(2.00L)}{0.08206 \frac{L \cdot atm}{mol \cdot K}(255K)}=0.03307\text{moles of Ne}$

Finally, convert the moles of neon into grams of neon using the molar mass.

$0.3307\text{ moles of Ne}(\frac{20.18\text{ g}}{1\text{ mole of Ne}})=0.667\text{ g}$

Make sure your answer has significant digits.

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Question

Suppose that hydrochloric acid and aluminum are reacted to produce hydrogen gas. This hydrogen gas is collected by displacement of water at $25^{o}C$ and a total pressure of . If the volume of the gas collected is , then how many moles of hydrogen were collected?

Note: The vapor pressure of water at $25^{o}C$ is .

Answer

For this question, we're told that a chemical reaction is producing hydrogen gas. In doing so, the gas is displacing a certain amount of water vapor. We're asked to determine the number of moles of gas produced in this process.

The amount of water displaced by the production of hydrogen gas will be the amount (volume) of hydrogen gas produced. To measure its pressure, we take the reading of the total pressure and then subtract from it the pressure of water vapor (since it contributes to the total pressure). Using this information, we can use the ideal gas equation to determine the number of moles of hydrogen gas.

$P_{total}=P_{water}+P_{H_{2}}$

$P_{H_{2}}=P_{total}-P_{water}$

$P_{H_{2}}=700:torr-23.8:torr=676.2:torr$

$676.2:torr\cdot \frac{1:atm}{760:torr}=0.890:atm$

Next, we can use the ideal gas equation to solve for our answer.

$n=\frac{PV}{RT}$

$n=\frac{(0.890:atm)(400\cdot 10^{-3}:L)}{(0.08206:\frac{L\cdot atm}{mol\cdot K})(298.15:K)}$

$n=1.46\cdot10^{-2}:moles\cdot\frac{1000:mmoles}{1:mole}=14.6:mmoles$

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Question

A 4.0 L container at 25 C is filled with 2.0 g neon and 8.0 g helium. What is the total pressure of the mixture?

Answer

To start, first recognize that this is an ideal gas law problem

$\mathrm{PV = \textit{n}RT}$

Rearranging the equation to solve for P gives us

$\mathrm{P = \frac{\mathit{n}RT}{V}}$

To solve the problem, we now need to find the total number of moles in the gaseous mixture _n_tot. To do this, convert the values for Ne and He to moles and add them together

$\textit{n}_{\textrm{tot}}=(2.0\textrm{ g Ne x }\frac{1\textrm{ mol Ne}}{20.2\textrm{ g Ne}})\textrm{ + }(8.0\textrm{ g He x }\frac{1\textrm{ mol He}}{4.003\textrm{ g He}})$

$n_{\textrm{tot}}=0.099\textrm{ mol Ne + }2.0\textrm{ mol He}=2.1\textrm{ mol gas}$

Now that we have _n_tot, plug known values into the ideal gas law equation. Remember to convert from Celsius to Kelvin!

$\textrm{P}=\frac{(2.0\textrm{ L)(0.0821}\frac{\textrm{L atm}}{\textrm{mol K}})(298.15\textrm{ K})}{4.0\textrm{ L}}=12\textrm{ atm}$

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Question

A 2.0 L container at 20 oC contains a gaseous mixture of $\textup{H}_2$ and He with a total pressure of 4 atm. The partial pressure exerted by He is 2.4 atm. How many moles of $\textup{H}_2$ are in the container?

Answer

This is another ideal gas law problem, but this time dealing with partial pressures. Total pressure of a mixture is the sum of the partial pressures of each component. So first calculate the partial pressure exerted by $\textup{H}_2$

$\textrm{P}_{\textrm{tot}}=\textrm{P}_{He}\textrm{ + P}_{H_{2}}$

$\textrm{4 atm = 2.4 atm + P}_{H_{2}}$

$\textrm{P}_{H_{2}}=4\textrm{ atm - 2.4 atm = 1.6 atm}$

Now that we have solved for the partial pressure of $\textup{H}_2$ , we can solve for moles of $\textup{H}_2$ by using the ideal gas law equation

$\textrm{P}_{H_{2}}=\frac{n_{H_{2}}\textrm{ x RT}}{\textrm{V}}$

Rearrange to solve for moles of $\textup{H}_2$

$n_{H_{2}}=\frac{\textrm{P}_{H_{2}}\textrm{ x V}}{\textrm{RT}}$

$n_{H_{2}}=\frac{(1.6\textrm{ atm)(2.0 L)}}{(0.0821 \frac{\textrm{L atm}}{\textrm{mol K}})(293.15\textrm{ K)}}=\textrm{0.13 atm}$

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