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Chapter 2: Problem 26

Under identical conditions of temperature, the density of a gas \(\mathrm{X}\) is two times that of gas \(\mathrm{Y}\) while molecular mass of gas \(Y\) is three times that of \(X\). The ratio of pressure of \(X\) and \(Y\) will be (a) \(2 / 3\) (b) \(3 / 2\) (c) 6 (d) 3

Short Answer

Expert verified
The ratio of pressure of gas X to gas Y is 6.

Step by step solution

01

Identify Known Variables

We have two gases, X and Y. The density of gas X is twice that of Y: \( \rho_X = 2 \rho_Y \). The molecular mass of Y is three times that of X: \( M_Y = 3 M_X \).
02

Use the Ideal Gas Law

Recall the ideal gas law in terms of density: \( P = \rho R T / M \), where \( P \) is pressure, \( \rho \) is density, \( R \) is the ideal gas constant, \( T \) is temperature, and \( M \) is molecular mass. This equation tells us that \( P \propto \frac{\rho}{M} \) when \( R \) and \( T \) are constant.
03

Set Up the Pressure Ratios

The pressure for gas X is \( P_X = \frac{\rho_X R T}{M_X} \) and for gas Y, \( P_Y = \frac{\rho_Y R T}{M_Y} \). We need the ratio of these pressures: \( \frac{P_X}{P_Y} = \frac{\rho_X / M_X}{\rho_Y / M_Y} \).
04

Substitute Known Values

Substitute the given relationships: \( \rho_X = 2 \rho_Y \) and \( M_Y = 3 M_X \). Substituting into the pressure ratio equation gives: \( \frac{P_X}{P_Y} = \frac{2 \rho_Y / M_X}{\rho_Y / (3 M_X)} \).
05

Simplify the Expression

Simplify the expression for the pressure ratio: \( \frac{P_X}{P_Y} = \frac{2}{1/3} = 6 \).
06

Final Answer

The ratio of the pressures \( P_X / P_Y \) is 6. Thus, the correct choice is (c) 6.

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Key Concepts

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

Gas Density
Gas density refers to the mass of gas per unit volume. It determines how compactly gas molecules are packed in a specific space. The density of a gas can vary depending on factors like molecular mass and temperature.
When talking about two different gases, like in our exercise, the ideal gas law helps to examine these relationships. For gas X, which has a density twice that of gas Y, this implies that the particles of gas X are more closely packed together.
Understanding how gas density works is key because it directly affects the pressure the gas exerts. Under consistent temperature conditions, a denser gas like gas X will impact the pressure calculations, playing a critical role in understanding pressure relations.
Molecular Mass
Molecular mass is essentially the mass of a single molecule of a substance, often measured in atomic mass units (amu). It plays a crucial role in understanding gas behavior and characteristics.
In the context of the ideal gas law, the molecular mass can affect the density and pressure of a gas. For instance, gas Y, which has three times the molecular mass of gas X, will have distinct properties affecting how it behaves under the same conditions. This leads to different expressions in the ideal gas law equation, where molecular mass inversely affects gas pressure when density is held constant.
  • Higher molecular mass usually means the particles are heavier, and this can influence how gases mix and react with each other.
  • Molecular mass helps in determining the molar volume of gases as well, impacting how gases are calculated in chemical reactions.
Pressure Ratio
The pressure ratio of two gases shows how their individual pressures compare under set conditions. It's influenced by factors like density and molecular mass, as discussed in the ideal gas law equation.
In our solution, knowing that the density of gas X is twice that of gas Y and the molecular mass of gas Y is three times that of gas X helps calculate this ratio.
By using the ideal gas law format, the expression \(\frac{P_X}{P_Y} = \frac{2 \rho_Y / M_X}{\rho_Y / (3 M_X)}\) simplifies to \(6\). Therefore, under the same conditions of temperature, the pressures of these two gases do not proportionally match their densities due to differences in their molecular masses.
Understanding pressure ratios can help predict how different gases behave in shared spaces, important for fields ranging from atmospheric studies to chemical engineering.
  • Changes in pressure ratios can signal shifts in physical conditions, which is critical in processes like gas expansion or compression.
  • Knowing pressure ratios helps in designing equipment and systems where gases are used, ensuring efficiency and safety.

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Most popular questions from this chapter

Two gases \(\mathrm{A}\) and \(\mathrm{B}\) present separately in two vessels \(X\) and \(Y\) at the same temperature with molecular weights \(\mathrm{M}\) and \(2 \mathrm{M}\) respectively are effused out. The orifice in vessel \(X\) is circular while that in \(Y\) is a square. If the radius of the circular orifice is equal to that of the length of the square orifice, the ration of rates of effusion of gas \(\mathrm{A}\) to that of gas \(\mathrm{B}\) is (a) \(\sqrt{2} \pi\) (b) \(\sqrt{\frac{\pi}{2}}\) (c) \(2 \pi\) (d) \(\sqrt{\frac{2}{\pi}}\)

\(5 \mathrm{ml}\) of \(\mathrm{N} \mathrm{HCl}, 20 \mathrm{ml}\) of \(\mathrm{N} / 2 \mathrm{H}_{2} \mathrm{SO}_{4}\) and \(30 \mathrm{ml}\) of \(\mathrm{N} / 3\) \(\mathrm{HNO}_{3}\) are mixed together and volume made one litre. The normality of the resulting solution is (a) \(\mathrm{N} / 5\) (b) \(\mathrm{N} / 10\) (c) \(\mathrm{N} / 20\) (d) N/40

\(\mathrm{X} \mathrm{ml}\) of \(\mathrm{H}_{2}\) gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is (a) 10 seconds : He (b) 20 seconds : \(\mathrm{O}_{2}\) (c) 25 seconds : \(\mathrm{CO}\) (d) 55 seconds : \(\mathrm{CO}_{2}\)

A mixture of \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{CH}_{4}\) exerts a pressure of \(320 \mathrm{~mm}\) \(\mathrm{Hg}\) at temperature \(\mathrm{T} \mathrm{K}\) in a \(\mathrm{V}\) litre flask. On complete combustion, gaseous mixture contains \(\mathrm{CO}_{2}\) only and exerts a pressure of \(448 \mathrm{~mm} \mathrm{Hg}\) under identical conditions. Hence mole fraction of \(\mathrm{C}_{3} \mathrm{H}_{8}\) in the mixture is (a) \(0.2\) (b) \(0.8\) (c) \(0.25\) (d) \(0.75\)

The density of a gaseous substance at 1 atm pressure and \(750 \mathrm{~K}\) is \(0.30 \mathrm{~g} / \mathrm{lt}\). If the molecular weight of the substance is 27 the forces existing among gas molecules is (a) Attractive (b) Repulsive (c) Both (a) and (b) (d) None of these
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