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Chapter 9: Problem 1312

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) \(8270 \mathrm{~K}\)

Short Answer

Expert verified
The temperature at which the rms speed of hydrogen molecules is equal to the escape velocity on Earth's surface is approximately \(10063 \, \text{K}\) (Option B).

Step by step solution

01

Understanding the given information and the formula for rms speed

We know that the rms speed (Vrms) of a gas molecule depends on temperature (T) and its molar mass (M). The formula for the rms speed is given by: Vrms = \(\sqrt{\dfrac{3RT}{M}}\) where R is the universal gas constant, which is \(8.314 \, \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}\), T is the temperature in Kelvin, and M is the molar mass in kg/mol. For our problem, we are given hydrogen as our gas. The molar mass of hydrogen is approximately 0.002 kg/mol.
02

Identifying the formula for escape velocity

The escape velocity (Vescape) on Earth's surface is given by: Vescape = \(\sqrt{\dfrac{2GM_e}{r_e}}\) where \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^{2} \cdot \text{kg}^{-2}\)), \(M_e\) is the Earth's mass (\(5.972 \times 10^{24} \, \text{kg}\)), and \(r_e\) is the Earth's radius (\(6.371 \times 10^{6} \, \text{m}\)).
03

Equating rms speed and escape velocity

To find the temperature at which the rms speed of hydrogen molecules is equal to the escape velocity on Earth, we will set the formulas for Vrms and Vescape equal to each other and solve for T: \(\sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{2GM_e}{r_e}}\)
04

Solving for temperature (T)

To solve for T, we first square both sides of the equation to remove the square root signs: \(\dfrac{3RT}{M} = \dfrac{2GM_e}{r_e}\) Now, we multiply both sides of the equation by M and divide by 3R: T = \(\dfrac{2GM_eM}{3Rr_e}\) Now, we will plug in the values for all the constants: T = \(\dfrac{2 \cdot (6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^{2} \cdot \text{kg}^{-2}) \cdot (5.972 \times 10^{24} \, \text{kg}) \cdot (0.002 \, \text{kg/mol})}{3 \cdot (8.314 \, \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}) \cdot (6.371 \times 10^{6} \, \text{m})}\)
05

Calculating the temperature (T)

Solving our equation for T, we get: T ≈ \(10063 \, \text{K}\) Thus, the temperature at which the rms speed of hydrogen molecules is equal to the escape velocity on Earth's surface is approximately 10063 K. The correct answer is (B).

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Most popular questions from this chapter

At what temperature the molecules of nitrogen will have the same rms. velocity as the molecules of Oxygen at \(127^{\circ} \mathrm{C}\). (A) \(273^{\circ} \mathrm{C}\) (B) \(350^{\circ} \mathrm{C}\) (C) \(77^{\circ} \mathrm{C}\) (D) \(457^{\circ} \mathrm{C}\)

The rms speed of gas molecules is given by (A) \(2.5 \sqrt{\left[M_{0} /(R T)\right]}\) (B) \(\left.2.5 \sqrt{[}(\mathrm{RT}) / \mathrm{M}_{0}\right]\) (C) \(\left.1.73 \sqrt{[}(\mathrm{RT}) / \mathrm{M}_{0}\right]\) (D) \(1.73 \sqrt{\left[M_{0} /(R T)\right]}\)

The rms velocity of gas molecules is \(300 \mathrm{~ms}^{-1}\). The rms velocity of molecules of gas with twice the molecular weight and half the absolute temperature is (A) \(300 \mathrm{~ms}^{-1}\) (B) \(150 \mathrm{~ms}^{-1}\) (C) \(600 \mathrm{~ms}^{-1}\) (D) \(75 \mathrm{~ms}^{-1}\)

The mean kinetic energy of a gas at \(300 \mathrm{~K}\) is \(100 \mathrm{~J}\). mean energy of the gas at \(450 \mathrm{~K}\) is equal to (A) \(100 \mathrm{~J}\) (B) \(150 \mathrm{~J}\) (C) \(3000 \mathrm{~J}\) (D) \(450 \mathrm{~J}\)

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) \([(1930) / \sqrt{3}] \mathrm{m} / \mathrm{s}\)
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