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Without looking at a table of values, which of the following gases would you expect to have the largest value of the van der Waals constant \(b: \mathrm{H}_{2}, \mathrm{~N}_{2}, \mathrm{CH}_{4}, \mathrm{C}_{2} \mathrm{H}_{6}\), or \(\mathrm{C}_{3} \mathrm{H}_{8}\) ?

Short Answer

Expert verified
The gas with the largest value of the van der Waals constant \(b\) would be C3H8 (Propane), as it has the largest molecular weight (\(44.096 \, amu\)) among the given gases. This is because molecular weight plays a significant role in determining the van der Waals constant \(b\).

Step by step solution

01

Analyzing the molecular weight

To determine which gas would have the largest value of the van der Waals constant b, we'll need to consider the molecular weight of each gas. Remember that the molecular weight is determined by the sum of the atomic weights of the atoms in the molecule.
02

Evaluating each gas's molecular weight

Let's evaluate the molecular weight of each gas: 1. H2 (Hydrogen gas): Its molecular weight is 2(1.008) = \(2.016 \, amu\) 2. N2 (Nitrogen gas): Its molecular weight is 2(14.007) = \(28.014 \, amu\) 3. CH4 (Methane): Its molecular weight is 12.011 + 4(1.008) = \(16.043 \, amu\) 4. C2H6 (Ethane): Its molecular weight is 2(12.011) + 6(1.008) = \(30.069 \, amu\) 5. C3H8 (Propane): Its molecular weight is 3(12.011) + 8(1.008) = \(44.096 \, amu\)
03

Identifying the gas with the largest molecular weight

Comparing the molecular weight of each gas, we can see that C3H8 (Propane) has the largest molecular weight of \(44.096 \, amu\).
04

Determining the gas with the largest value of the van der Waals constant b

Since molecular weight plays a significant role in determining the van der Waals constant b, we can conclude that C3H8 (Propane) would have the largest value of the van der Waals constant b among the given gases due to its largest molecular weight.

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Key Concepts

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

Molecular Weight
Molecular weight is a critical factor when assessing the properties of a gas, as seen in our problem examining the van der Waals constant. Molecular weight is calculated by summing up the atomic weights of the atoms present in a molecule. For each gas:
  • Hydrogen gas, \( \mathrm{H}_2 \), with a molecular weight of approximately \(2.016 \, \text{amu} \).
  • Nitrogen gas, \( \mathrm{N}_2 \), weighs in at \(28.014 \, \text{amu} \).
  • Methane, \( \mathrm{CH}_4 \), has a molecular weight of \(16.043 \, \text{amu} \).
  • Ethane, \( \mathrm{C}_2\mathrm{H}_6 \), comes in at \(30.069 \, \text{amu} \).
  • Propane, \( \mathrm{C}_3\mathrm{H}_8 \), possesses the largest molecular weight at \(44.096 \, \text{amu} \).
These weights are rounded based on the atomic masses of hydrogen, carbon, and nitrogen. Because molecular weight impacts volume and interaction characteristics of gases, it is a foundational metric in predicting gas behavior, including influences on the van der Waals constants.
Gas Properties
Gas properties are vital in understanding how gases behave under different conditions. Gases are composed of many molecules that move freely and rapidly. The behaviors of gases can be influenced by several key properties:
  • Volume: Gases occupy space and the volume affects the space that gases fill.
  • Pressure: Gas molecules collide with container walls, creating pressure.
  • Temperature: Increasing temperature raises the kinetic energy of gas molecules.
  • Molecular Interactions: At higher pressures or lower temperatures, gases exhibit non-ideal behavior due to intermolecular forces.
  • Molecular Size: Larger molecules generally occupy more space and experience stronger interactions.
Understanding these properties helps evaluate equations of states, like the ideal gas law and van der Waals equation, predicting gas behavior under various circumstances.
van der Waals Equation
The van der Waals equation is a mathematical formula used to describe the behavior of real gases, providing a way to account for deviations from the ideal gas law. The equation incorporates two critical correction factors:\[\left( P + \frac{a}{V_m^2} \right)(V_m-b) = RT\]Where:
  • \( P \) stands for pressure of the gas.
  • \( V_m \) represents the molar volume.
  • \( a \) corrects for intermolecular forces.
  • \( b \) corrects for the finite size of molecules (related to molecular volume).
  • \( R \) is the ideal gas constant.
  • \( T \) is the temperature.
The constant \( b \) particularly reflects the volume occupied by gas molecules. Larger molecules with higher molecular weights tend to have larger values of \( b \), as seen in the exercise where propane has the largest \( b \). This means they take up more space in the gas phase. The van der Waals equation helps predict how real gases deviate from ideal behavior by accounting for the particle volume and intermolecular forces.

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Most popular questions from this chapter

Calculate the pressure exerted by \(0.5000 \mathrm{~mol} \mathrm{~N}_{2}\) in a \(10.000-\mathrm{L}\) container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results. d. Compare the results with those in Exercise 107 .

Trace organic compounds in the atmosphere are first concentrated and then measured by gas chromatography. In the concentration step, several liters of air are pumped through a tube containing a porous substance that traps organic compounds. The tube is then connected to a gas chromatograph and heated to release the trapped compounds. The organic compounds are separated in the column and the amounts are measured. In an analysis for benzene and toluene in air, a \(3.00-\mathrm{L}\) sample of air at 748 torr and \(23^{\circ} \mathrm{C}\) was passed through the trap. The gas chromatography analysis showed that this air sample contained \(89.6\) ng benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) and \(153 \mathrm{ng}\) toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right) .\) Calculate the mixing ratio (see Exercise 121 ) and number of molecules per cubic centimeter for both benzene and toluene.

A mixture of chromium and zinc weighing \(0.362 \mathrm{~g}\) was reacted with an excess of hydrochloric acid. After all the metals in the mixture reacted, \(225 \mathrm{~mL}\) dry hydrogen gas was collected at \(27^{\circ} \mathrm{C}\) and 750 . torr. Determine the mass percent of \(\mathrm{Zn}\) in the metal sample. [Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas; chromium reacts with hydrochloric acid to produce chromium(III) chloride and hydrogen gas.]

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g)\), ammonia, \(\mathrm{NH}_{3}\left(\mathrm{~g}\right.\) ), and oxygen, \(\mathrm{O}_{2}(\mathrm{~g})\), at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of \(\mathrm{HCN}(\mathrm{g})\) can be obtained from the reaction of \(20.0 \mathrm{~L} \mathrm{CH}_{4}(g), 20.0 \mathrm{~L} \mathrm{NH}_{3}(g)\), and \(20.0 \mathrm{~L} \mathrm{O}_{2}(g) ?\) The volumes of all gases are measured at the same temperature and pressure.

As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant pressure and temperature, the volume of the product gases collected is twice the volume of \(\mathrm{NH}_{3}\) reacted. Explain. As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant volume and temperature, the total pressure increases by some factor. Why the increase in pressure and by what factor does the total pressure increase when reactants are completely converted into products? How do the partial pressures of the product gases compare to each other and to the initial pressure of \(\mathrm{NH}_{3}\) ?

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