A mixture of \(1.00 \mathrm{~g} \mathrm{H}_{2}\) and \(1.00 \mathrm{~g} \mathrm{He}\) is placed in a \(1.00-\mathrm{L}\) container at \(27^{\circ} \mathrm{C}\). Calculate the partial pressure of each gas and the total pressure.

Short Answer

Expert verified
In summary, the partial pressure of hydrogen (H2) is 12.32 atm, the partial pressure of helium (He) is 6.16 atm, and the total pressure of the mixture is 18.48 atm.

Step by step solution

01

Convert grams to moles for both gases

To convert grams to moles, we can use the given mass of each gas and their respective molar mass. The molar mass of hydrogen (H2 molecule) is 2 g/mol, and the molar mass of helium (He atom) is 4 g/mol. For hydrogen (H2): \(n_{H2} = \frac{mass_{H2}}{molar\_mass_{H2}} = \frac{1.00 g}{2 g/mol} = 0.50 mol\) For helium (He): \(n_{He} = \frac{mass_{He}}{molar\_mass_{He}} = \frac{1.00 g}{4 g/mol} = 0.25 mol\)
02

Convert Celsius to Kelvin

In order to use the ideal gas law, we need to express the temperature in Kelvin. The conversion from Celsius to Kelvin is achieved by adding 273.15 to the given temperature in Celsius. \(T = 27^{\circ}C + 273.15 = 300.15K\)
03

Use the Ideal Gas Law for partial pressures

The Ideal Gas Law is expressed as \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin. We need to find the partial pressure, so we need to rearrange the formula as follows: \(P = \frac{nRT}{V}\) For hydrogen (H2): \(P_{H2} = \frac{n_{H2}RT}{V} = \frac{0.50 mol \times 0.0821 \frac{L \cdot atm}{mol \cdot K} \times 300.15 K}{1.00 L} = 12.32 atm\) For helium (He): \(P_{He} = \frac{n_{He}RT}{V} = \frac{0.25 mol \times 0.0821 \frac{L \cdot atm}{mol \cdot K} \times 300.15 K}{1.00 L} = 6.16 atm\)
04

Calculate the total pressure

The total pressure of the gas mixture can be obtained by adding the partial pressures of H2 and He gases. Total pressure = \(P_{H2} + P_{He} = 12.32 atm + 6.16 atm = 18.48 atm\) In summary, the partial pressure of hydrogen (H2) is 12.32 atm, the partial pressure of helium (He) is 6.16 atm, and the total pressure of the mixture is 18.48 atm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Ideal Gas Law
The Ideal Gas Law is a fundamental principle in chemistry that relates the pressure, volume, temperature, and the number of moles of an ideal gas. It is represented by the equation \(PV=nRT\), where \(P\) stands for pressure, \(V\) represents volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.

The gas constant \(R\) has different values depending on the units used, but for the purpose of calculating pressure in atmospheres (atm) when volume is in liters (L), moles (mol), and temperature in Kelvin (K), the value is typically \(0.0821 L\cdot atm/mol\cdot K\). Temperature must always be in Kelvin for the Ideal Gas Law to be accurate, as the Kelvin scale is an absolute scale with \(0\) being absolute zero.

This law is especially useful because it allows us to solve for any single variable if we know the other three. For example, in the exercise, we use the Ideal Gas Law to find out the partial pressure exerted by each gas in a mixture by rearranging the formula to \(P = \frac{nRT}{V}\).
Molar Mass Conversion in Gas Calculations
Molar mass conversion is essential when working with gases to convert between mass and moles, as calculations involving gases often use moles rather than grams. The molar mass is defined as the mass of one mole of a substance, and it is usually expressed in grams per mole (g/mol).

For instance, in the given problem, the molar mass of hydrogen (H2) is \(2\ g/mol\), while that of helium (He) is \(4\ g/mol\). To find the number of moles (\(n\)) from the given mass, you divide the mass of the gas by its molar mass (\(n = \frac{mass}{molar\_mass}\)). This step is crucial because the Ideal Gas Law requires the amount of gas to be in moles.

Once the number of moles is determined, as shown in the exercise, one can proceed to use the Ideal Gas Law and other related calculations. Remember that precise molar mass values are necessary for accurate results.
Calculating Gas Mixture Pressure
When dealing with a mixture of gases, like the one described in the exercise, the total pressure can be perceived as the sum of the partial pressures of each individual gas in the mixture. This concept is described by Dalton's Law of Partial Pressures.

Dalton's Law states that for a mixture of non-reacting gases, the total pressure exerted by the mixture is equal to the sum of the partial pressures of individual gases. Each gas in the mixture exerts pressure as if it were the only gas present, and can be calculated using the Ideal Gas Law.

In our example, after finding the partial pressures of hydrogen (H2) and helium (He) separately, we simply add them to find the total pressure of the gas mixture (\(P_{total} = P_{H2} + P_{He}\)). Knowing how to calculate the partial pressures and the total pressure of a mixture is vital in understanding how gases behave in different environments and is also used in various scientific and industrial applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider separate \(1.0\) - \(\mathrm{L}\) samples of \(\mathrm{He}(g)\) and \(\mathrm{UF}_{6}(g)\), both at \(1.00\) atm and containing the same number of moles. What ratio of temperatures for the two samples would produce the same root mean square velocity?

Silane, \(\mathrm{SiH}_{4}\), is the silicon analogue of methane, \(\mathrm{CH}_{4}\). It is prepared industrially according to the following equations: $$ \begin{aligned} \mathrm{Si}(s)+3 \mathrm{HCl}(g) & \longrightarrow \operatorname{HSiCl}_{3}(l)+\mathrm{H}_{2}(g) \\ 4 \mathrm{HSiCl}_{3}(l) & \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l) \end{aligned} $$ a. If \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}(d=1.34 \mathrm{~g} / \mathrm{mL})\) is isolated when \(15.0 \mathrm{~L}\) \(\mathrm{HCl}\) at \(10.0 \mathrm{~atm}\) and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of \(\mathrm{HSiCl}_{3}\) ? b. When \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at \(10.0\) atm and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is \(93.1 \%\) ?

Consider two gases, A and B, each in a 1.0-L container with both gases at the same temperature and pressure. The mass of gas \(\mathrm{A}\) in the container is \(0.34 \mathrm{~g}\) and the mass of gas \(\mathrm{B}\) in the container is \(0.48 \mathrm{~g}\). a. Which gas sample has the most molecules present? Explain. b. Which gas sample has the largest average kinetic energy? Explain. c. Which gas sample has the fastest average velocity? Explain. d. How can the pressure in the two containers be equal to each other since the larger gas B molecules collide with the container walls more forcefully?

You have a helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is \(79.0 \%\) nitrogen and \(21.0 \%\) oxygen by volume. The "lift" of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C}\) ? Explain. b. Calculate the temperature of the air required for the hotair balloon to provide the same lift as the helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). Assume atmospheric conditions are \(1.00\) atm and \(25^{\circ} \mathrm{C}\).

A hot-air balloon is filled with air to a volume of \(4.00 \times\) \(10^{3} \mathrm{~m}^{3}\) at 745 torr and \(21^{\circ} \mathrm{C}\). The air in the balloon is then heated to \(62^{\circ} \mathrm{C}\), causing the balloon to expand to a volume of \(4.20 \times 10^{3} \mathrm{~m}^{3}\). What is the ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon? (Hint: Openings in the balloon allow air to flow in and out. Thus the pressure in the balloon is always the same as that of the atmosphere.)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free