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Chapter 9: Problem 1281

If the ratio of vapor density for hydrogen and oxygen is \([1 /(16)]\), then under constant pressure the ratio of their \(\mathrm{rms}\) velocities will be (A) \(4: 1\) (B) \(1: 16\) (C) \(16: 1\) (D) \(1: 4\)

Short Answer

Expert verified
The ratio of the RMS velocities of hydrogen and oxygen under constant pressure is 4:1. The correct answer is (A) \(4: 1\).

Step by step solution

01

Find the relationship between vapor density, molecular mass and RMS velocity

The vapor density (ρ) of a gas is directly proportional to its molecular mass (M). Mathematically, this can be expressed as: ρ ∝ M The RMS velocity (v) of a gas can be calculated using the formula: \(v = \sqrt{\frac{3kT}{m}}\) Where: - v is the RMS velocity - k is the Boltzmann constant (1.38 x 10^-23 J/K) - T is the temperature in Kelvin - m is the mass of the gas molecules Since the pressure is constant, temperature (T) and Boltzmann's constant (k) can be considered constant. Therefore, the relationship between RMS velocity and molecular mass can be written as: v ∝ 1 / √M
02

Use the given ratio of vapor densities to find the ratio of molecular masses

The given ratio of vapor densities for hydrogen and oxygen is [1 / 16]. Using the relationship between vapor density and molecular mass, we can write the ratio of molecular masses as follows: M_H / M_O = 1 / 16
03

Use the ratio of molecular masses to find the ratio of RMS velocities

Now, we can use the relationship between RMS velocity and molecular mass along with the ratio of molecular masses to find the ratio of RMS velocities: v_H / v_O = √(M_O / M_H) Substituting the ratio of molecular masses, we get: v_H / v_O = √(16 / 1) v_H / v_O = 4 / 1 Thus, the ratio of the RMS velocities of hydrogen and oxygen under constant pressure is 4:1. The correct answer is (A) \(4: 1\).

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Most popular questions from this chapter

\(\mathrm{O}_{2}\) gas is filled in a vessel. If pressure is double, temperature becomes four times, how many times its density will become. (A) 4 (B) \((1 / 4)\) (C) 2 (D) \((1 / 2)\)

A gas at \(27^{\circ} \mathrm{C}\) temperature and 30 atmospheric pressure $1 \mathrm{~s}$ allowed to expand to the atmospheric pressure if the volume becomes two times its initial volume, then the final temperature becomes (B) \(-173^{\circ} \mathrm{C}\) (A) \(273^{\circ} \mathrm{C}\) (C) \(173^{\circ} \mathrm{C}\) (D) \(100^{\circ} \mathrm{C}\)

The relation between the gas pressure \(\mathrm{P}\) and average kinetic energy per unit volume \(E\) is (A) \(\mathrm{P}=(2 / 3) \mathrm{E}\) (B) \(P=(3 / 2) E\) (C) \(\mathrm{P}=\mathrm{E}\) (D) \(\mathrm{P}=(\mathrm{E} / 2)\)

The volume of a gas at \(20 \mathrm{C}\) is \(200 \mathrm{ml}\). If the temperature is reduced to \(-20^{\circ} \mathrm{C}\) at constant pressure, its volume will be. (A) \(172.6 \mathrm{~m} 1\) (B) \(17.26 \mathrm{ml}\) (C) \(19.27 \mathrm{ml}\) (D) \(192.7 \mathrm{ml}\)

At what temperature, pressure remaining unchanged, will the rms velocity of a gas be half its value at \(0^{\circ} \mathrm{C}\) ? (A) \(204.75 \mathrm{~K}\) (B) \(204.75^{\circ} \mathrm{C}\) (C) \(-204.75 \mathrm{~K}\) (D) \(-204.75^{\circ} \mathrm{C}\)
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