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Chapter 9: Problem 1298

The ratio of the vapor densities of two gases at a given temperature is \(9: 8\), The ratio of the rms velocities of their molecule is (A) \(3: 2 \sqrt{2}\) (B) \(2 \sqrt{2}: 3\) (C) \(9: 8\) (D) \(8: 9\)

Short Answer

Expert verified
The ratio of the rms velocities of the two gases is \(\dfrac{v_{1_{rms}}}{v_{2_{rms}}} = \dfrac{2\sqrt{2}}{3}\), so the correct answer is (B) \(2 \sqrt{2}: 3\).

Step by step solution

01

Set up the ratios of vapor densities and rms velocities

We are given the ratio of vapor densities as \[\dfrac{\rho_1}{\rho_2} = \dfrac{9}{8}\] Our aim is to find the ratio of the rms velocities of the gases: \[\dfrac{v_{1_{rms}}}{v_{2_{rms}}}\]
02

Use the formulas to express the given ratio

Express the given ratio of vapor densities with the formula: \[\dfrac{\dfrac{P_1M_1}{RT_1}}{\dfrac{P_2M_2}{RT_2}} = \dfrac{9}{8}\] Since the temperatures and pressures are given to be equal for both gases, we can rewrite the equation as: \[\dfrac{M_1}{M_2} = \dfrac{9}{8}\]
03

Use the rms velocity formula to express the desired ratio

Express the desired ratio of rms velocities using the rms velocity formula: \[\dfrac{\sqrt{\dfrac{3R T_1}{M_1}}}{\sqrt{\dfrac{3R T_2}{M_2}}} = \dfrac{v_{1_{rms}}}{v_{2_{rms}}}\] Since the temperatures are equal, we can rewrite the equation as: \[\dfrac{\sqrt{\dfrac{M_2}{M_1}}}{\sqrt{\dfrac{M_1}{M_2}}} = \dfrac{v_{1_{rms}}}{v_{2_{rms}}}\]
04

Combine the equations and solve for the desired ratio

Using the ratio of molar masses from Step 2, we have: \[\dfrac{\sqrt{\dfrac{8}{9}}}{\sqrt{\dfrac{9}{8}}} = \dfrac{v_{1_{rms}}}{v_{2_{rms}}}\] Simplifying the expression gives: \[\dfrac{\sqrt{8}}{\sqrt{9}} = \dfrac{2\sqrt{2}}{3}\] So, the ratio of the rms velocities of the two gases is: \[\dfrac{v_{1_{rms}}}{v_{2_{rms}}} = \dfrac{2\sqrt{2}}{3}\] Therefore, the correct answer is (B) \(2 \sqrt{2}: 3\).

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Key Concepts

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

Vapor Density
Vapor density refers to the density of a vapor in comparison to air. It is an important concept when dealing with gases, as it helps us understand how different gases behave under similar conditions. The vapor density of a gas is directly related to its molar mass and can be calculated by comparing it to a reference gas, often air. To calculate the vapor density, we use the equation:
  • \(\text{Vapor Density} = \frac{\text{Molar Mass of Gas}}{\text{Molar Mass of Air}}\)
This equation tells us how much heavier or lighter the gas is compared to air. In the context of the exercise, the ratio of vapor densities \(\frac{9}{8}\) indicates that one gas is 9/8 times denser than the other, reflecting differences in molar mass between the two gases.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation that describes the behavior of ideal gases. It is represented as:\[ PV = nRT \] where:
  • \(P\) is the pressure of the gas,
  • \(V\) is the volume,
  • \(n\) is the number of moles,
  • \(R\) is the universal gas constant, and
  • \(T\) is the temperature in Kelvin.
This law combines several key relationships: Boyle鈥檚 Law, Charles鈥檚 Law, and Avogadro鈥檚 Law. In the context of the given problem, the Ideal Gas Law is implicit when we assume equal pressure and temperature for both gases, simplifying the calculation of vapor densities and molar masses. This simplification allows us to focus on the relationships between molar mass and other properties, like vapor density.
Molar Mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It plays a crucial role in gas calculations because it directly affects properties such as vapor density and root mean square (RMS) velocity.To calculate molar mass:
  • Use the periodic table to find the atomic mass for each element in a compound.
  • Add up the atomic masses based on the chemical formula of the compound.
In the exercise, we determine the ratio of molar masses \(\frac{M_1}{M_2} = \frac{9}{8}\) from the given vapor density ratio. This information is key to finding the RMS velocity ratio since the molecular weight affects the velocity of gas molecules. Generally, lighter gases have higher RMS velocities.
Kinetic Theory of Gases
The kinetic theory of gases provides a microscopic explanation of gas behavior, linking macroscopic properties such as pressure and temperature to the motion of gas molecules.Key points of the kinetic theory include:
  • Gases consist of tiny particles in constant random motion.
  • The volume of the gas particles is negligible compared to the volume of their container.
  • Gas particles do not exert forces on each other except during collisions.
  • The average kinetic energy of gas particles is directly proportional to the temperature in Kelvin.
The concept of root mean square (RMS) velocity emerges from this theory. RMS velocity is the measure of the average speed of gas particles in a system. It can be calculated using the formula:\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]where \(M\) represents molar mass. The problem utilizes this kinetic concept to relate molar mass ratios to RMS velocity ratios, demonstrating how temperature and molar mass influence molecular speed.

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Most popular questions from this chapter

The temperature of a gas at pressure \(\mathrm{P}\) and volume \(\mathrm{V}\) is \(27^{\circ} \mathrm{C}\) Keeping its volume constant if its temperature is raised to \(927^{\circ} \mathrm{C}\), then its pressure will be (A) \(3 \mathrm{P}\) (B) \(2 \mathrm{P}\) (C) \(4 \mathrm{P}\) (D) \(6 \mathrm{P}\)

The average translational kinetic energy of \(\mathrm{O}_{2}\) (molar mass 32 ) molecules at a particular temperature \(0.068 \mathrm{eV}\). The translational kinetic energy of \(\mathrm{N}_{2}\) (molar mass 28 ) molecules in \(\mathrm{eV}\) at the same temperature is (A) \(0.003 \mathrm{eV}\) (B) \(0.068 \mathrm{eV}\) (C) \(0.056 \mathrm{eV}\) (D) \(0.678 \mathrm{eV}\)

A cylinder contains \(10 \mathrm{~kg}\) of gas at pressure of \(10^{\prime} \mathrm{N} / \mathrm{m}^{2}\). The quantity of gas taken out of the cylinder, if final pressure is \(2.5 \times 10^{6} \mathrm{Nm}^{-2}\). will be (temperature of gas is constant) (A) \(5.2 \mathrm{~kg}\) (B) \(3.7 \mathrm{~kg}\) (C) \(7.5 \mathrm{~kg}\) (D) \(1 \mathrm{~kg}\)

To double the volume of a given mass at an ideal gas at \(27^{\circ} \mathrm{C}\) keeping the pressure constant one must raise the temperature in degree centigrade (A) \(54^{\circ}\) (B) \(600^{\circ}\) (C) \(327^{\circ}\) (D) \(270^{\circ}\)

The volume of a gas at pressure \(21 \times 10^{4} \mathrm{Nm}^{-2}\) and temperature \(27^{\circ} \mathrm{C}\) is 83 Liters. If \(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\). Then the quantity of gas in \(\mathrm{g}\) -mole will be (A) 42 (B) 7 (C) 14 (D) 15
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