/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 A piece of solid carbon dioxide,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A piece of solid carbon dioxide, with a mass of \(7.8 \mathrm{~g}\), is placed in a 4.0-L otherwise empty container at \(27^{\circ} \mathrm{C}\). What is the pressure in the container after all the carbon dioxide vaporizes? If \(7.8 \mathrm{~g}\) solid carbon dioxide were placed in the same container but it already contained air at 740 torr, what would be the partial pressure of carbon dioxide and the total pressure in the container after the carbon dioxide vaporizes?

Short Answer

Expert verified
In the first scenario (empty container), the pressure after all the COâ‚‚ vaporizes is 1.093 atm. In the second scenario (container with air at 740 torr), the partial pressure of COâ‚‚ is 1.093 atm, and the total pressure after the COâ‚‚ vaporizes is 2.067 atm.

Step by step solution

01

Convert mass of carbon dioxide to moles

First, we need to convert the 7.8 g of carbon dioxide to moles. We can do this using the molar mass of carbon dioxide, which is 44.01 g/mol. moles of COâ‚‚ = mass of COâ‚‚ / molar mass of COâ‚‚ moles of COâ‚‚ = 7.8 g / 44.01 g/mol = 0.177 moles
02

Convert temperature to Kelvin

Next, we need to convert the given temperature from Celsius to Kelvin using the following formula: Temperature in Kelvin (K) = Temperature in Celsius (°C) + 273.15 Temperature in Kelvin = 27 °C + 273.15 = 300.15 K
03

Calculate pressure in the first scenario (empty container)

Now we can use the Ideal Gas Law (PV = nRT) to find the pressure after all the CO₂ vaporizes in the empty container. P = (nRT) / V P = (0.177 moles × 0.0821 L×atm/mol×K × 300.15 K) / 4.0 L P = 1.093 atm Thus, the pressure in the container after all the CO₂ vaporizes is 1.093 atm.
04

Calculate the pressure of air in the second scenario

In the second scenario, we are given that the container already contains air at a pressure of 740 torr. We will first need to convert this pressure to atm, using the conversion factor of 1 atm = 760 torr. Pressure of air in atm = Pressure of air in torr / 760 torr Pressure of air in atm = 740 torr / 760 torr = 0.974 atm
05

Calculate the total pressure in the second scenario

To find the total pressure in the container after the COâ‚‚ vaporizes, we will add the partial pressure of COâ‚‚ (found in Step 3) to the pressure of air (found in Step 4). Total pressure = Partial pressure of COâ‚‚ + Pressure of air Total pressure = 1.093 atm + 0.974 atm = 2.067 atm
06

Find the partial pressures of COâ‚‚ and air in the second scenario

Finally, we want to find both the partial pressure of COâ‚‚ and the total pressure in the container after the COâ‚‚ vaporizes. The partial pressure of COâ‚‚ is the same as the pressure in the first scenario, which we found in Step 3. Therefore, the partial pressure of COâ‚‚ is 1.093 atm. The total pressure in the container was found in Step 5. So, the total pressure is 2.067 atm. In conclusion, - In the first scenario where the container is empty, the pressure after all the COâ‚‚ vaporizes is 1.093 atm. - In the second scenario where the container already contains air, the partial pressure of COâ‚‚ is 1.093 atm, and the total pressure after the COâ‚‚ vaporizes is 2.067 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 91Ó°ÊÓ!

Key Concepts

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

Partial Pressure
Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. It is important to understand that each gas in a container exerts pressure independently of other gases present. This means that the partial pressure of a gas is not affected by the presence of other gases.

To calculate the partial pressure of a gas, we use the ideal gas law, which is expressed as:\[\text{PV} = \text{nRT}\]where:
  • \(P\) is the pressure in atm.
  • \(V\) is the volume in liters.
  • \(n\) is the number of moles of the gas.
  • \(R\) is the ideal gas constant, typically 0.0821 L atm/mol K.
  • \(T\) is the temperature in Kelvin.
The partial pressure can be found by solving the ideal gas equation for \(P\), which involves determining the number of moles \(n\) of the specific gas. When multiple gases are present, their partial pressures sum up to give the total pressure in the container.
Molar Mass
Molar mass is a fundamental concept in chemistry, representing the mass of one mole of a substance, expressed in grams per mole (g/mol). For carbon dioxide (COâ‚‚), the molar mass is calculated by adding the atomic masses of its constituent elements: carbon (C) and oxygen (O).

- Carbon has an atomic mass of approximately 12.01 g/mol.- Oxygen has an atomic mass of approximately 16.00 g/mol.

Since a molecule of COâ‚‚ contains one carbon atom and two oxygen atoms, its molar mass is calculated as:\[\text{Molar Mass of } CO_2 = 12.01 + (2 \times 16.00) = 44.01 \text{ g/mol}\]Molar mass allows us to convert between the mass of a substance and the number of moles, which is crucial when using the ideal gas law. For instance, in determining the moles of COâ‚‚ from a mass of 7.8 g, the formula is:\[\text{moles of CO}_2 = \frac{\text{mass of CO}_2}{\text{molar mass of CO}_2}\]Knowing the moles is essential for understanding how much gas is present and predicting its behavior under various conditions.
Temperature Conversion
When working with gas laws, it is essential to use temperatures in Kelvin rather than Celsius or Fahrenheit. Kelvin is the SI unit for temperature, and it's necessary because the Ideal Gas Law requires an absolute temperature scale.

To convert from Celsius to Kelvin, use the formula:\[\text{Temperature in Kelvin (K)} = \text{Temperature in Celsius (°C)} + 273.15\]This conversion ensures that the temperature is always positive, which is vital for calculations involving gases, as a zero or negative temperature in Celsius would not make sense in the Ideal Gas Law.

For example, the original problem set the temperature at 27°C, which converts to:\[27°C + 273.15 = 300.15 \text{ K}\]Kelvin is a reflection of the kinetic energy of the gas particles. Hence, accurate temperature conversion is critical for determining properties like pressure or volume.
Gas Laws
Gas laws are a set of rules that describe the behavior of gases under various conditions. The Ideal Gas Law is a central piece, expressed as: \[PV = nRT\]This law combines several simpler laws:
  • Boyle's Law, which says that pressure and volume are inversely proportional at a constant temperature.
  • Charles's Law, which states that volume and temperature are directly proportional at a constant pressure.
  • Avogadro's Law, which indicates that volume and the number of moles are directly proportional when pressure and temperature are constant.
Each of these laws addresses how gas properties change and interact. Understanding these concepts helps predict how a gas behaves in different scenarios, such as changes in temperature or when adding more gas to a container.

The Ideal Gas Law, specifically, is useful for calculating unknown variables like pressure, as seen in the carbon dioxide problem, where we determined pressure after vaporization based on fixed volume and temperature conditions.

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

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 \(25^{\circ} \mathrm{C}\).

An 11.2-L sample of gas is determined to contain \(0.50\) mole of \(\mathrm{N}_{2}\). At the same temperature and pressure, how many moles of gas would there be in a 20.-L sample?

A balloon is filled to a volume of \(7.00 \times 10^{2} \mathrm{~mL}\) at a temperature of \(20.0^{\circ} \mathrm{C}\). The balloon is then cooled at constant pressure to a temperature of \(1.00 \times 10^{2} \mathrm{~K}\). What is the final volume of the balloon?

Explain the following seeming contradiction: You have two gases, \(A\) and \(B\), in two separate containers of equal volume and at equal pressure and temperature. Therefore, you must have the same number of moles of each gas. Because the two temperatures are equal, the average kinetic energies of the two samples are equal. Therefore, since the energy given such a system will be converted to translational motion (that is, move the molecules), the root mean square velocities of the two are equal, and thus the particles in each sample move, on average, with the same relative speed. Since \(A\) and \(B\) are different gases, they each must have a different molar mass. If \(A\) has a higher molar mass than \(B\), the particles of \(A\) must be hitting the sides of the container with more force. Thus the pressure in the container of gas \(A\) must be higher than that in the container with gas \(B\). However, one of our initial assumptions was that the pressures were equal.

If you release a helium balloon, it soars upward and eventually pops. Explain this behavior.

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.