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:
- \(\text{Vapor Density} = \frac{\text{Molar Mass of Gas}}{\text{Molar Mass of Air}}\)
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.
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.
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.