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Chapter 9: Problem 1308

The root mean square speed of hydrogen molecules at \(300 \mathrm{~K}\) is $1930 \mathrm{~m} / \mathrm{s}$. Then the root mean square speed of Oxygen molecules at \(900 \mathrm{~K}\) will be (A) \(836 \mathrm{~m} / \mathrm{s}\) (B) \(643 \mathrm{~m} / \mathrm{s}\) (C) \(1930 \sqrt{3} \mathrm{~m} / \mathrm{s}\) (D) \([(1930) / \sqrt{3}] \mathrm{m} / \mathrm{s}\)

Short Answer

Expert verified
The short answer is: The root mean square speed of Oxygen molecules at \(900 \mathrm{~K}\) will be (B) \(643 \mathrm{~m} / \mathrm{s}\).

Step by step solution

01

Calculate the molar mass of hydrogen and Oxygen molecules.

First, we need to know the molar mass of hydrogen (H) and oxygen (O) molecules: - Hydrogen molecule (H2): 1g/mol × 2 = 2 g/mol = \(2×10^{-3}\) kg/mol - Oxygen molecule (O2): 16g/mol × 2 = 32 g/mol = \(32×10^{-3}\) kg/mol
02

Find the ratio of the root mean square speeds of hydrogen and oxygen molecules.

According to the root mean square speed formula: $$\frac{v_{rms_{H}}}{v_{rms_{O}}} = \sqrt{\frac{3R T_{H}}{M_{H}} \times \frac{M_{O}}{3 R T_{O}}} = \sqrt{\frac{M_O T_H}{M_H T_O}}$$ We can now substitute the given values and molar masses into the equation: $$\frac{1930}{v_{rms_{O}}} = \sqrt{\frac{(32×10^{-3})(300)}{(2×10^{-3})(900)}}$$
03

Calculate the root mean square speed of oxygen molecules at 900 K.

Now we need to solve for \(v_{rms_O}\): $$v_{rms_{O}}= \frac{1930}{\sqrt{\frac{(32×10^{-3})(300)}{(2×10^{-3})(900)}}}$$ Calculate the value: $$v_{rms_{O}}= 643 \mathrm{~m} / \mathrm{s}$$ So the correct answer is (B) \(643 \mathrm{~m} / \mathrm{s}\).

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Most popular questions from this chapter

The ratio of mean kinetic energy of hydrogen and oxygen at a given temperature is (A) \(1: 8\) (B) \(1: 4\) (C) \(1: 16\) (D) \(1: 1\)

Pressure of an ideal gas is increased by keeping temperature constant what is the effect on kinetic energy of molecules. (A) Decrease (B) Increase (C) No change (D) Can't be determined

The root mean square velocity of the molecules in a sample of helium is $(5 / 7)^{\text {th }}$ that of the molecules in a sample of hydrogen. If the temperature of hydrogen sample is \(0^{\circ} \mathrm{C}\), then the temperature of the helium sample is about (A) \(273^{\circ} \mathrm{C}\) (B) \(0^{\circ} \mathrm{C}\) (C) \(0^{\circ} \mathrm{K}\) (D) \(100^{\circ} \mathrm{C}\)

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) \(1 \mathrm{~kg}\)

The temperature at which the rms speed of hydrogen molecules is equal to escape velocity on earth surface will be (A) \(5030 \mathrm{~K}\) (B) \(10063 \mathrm{~K}\) (C) \(1060 \mathrm{~K}\) D) \(8270 \mathrm{~K}\)
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