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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

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 root mean square velocity of a gas molecule of mass \(\mathrm{m}\) at a given temperature is proportional to (A) \(\mathrm{m}^{0}\) (B) \(\mathrm{m}^{-1 / 2}\) (C) \(\mathrm{m}^{1 / 2}\) (D) \(\mathrm{m}\)

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) \(-204.75^{\circ} \mathrm{C}\)

A diatomic molecule has how many degrees of freedom (For rigid rotator) (A) 4 (B) 3 (C) 6 (D) 5

A vessel contains 1 mole of \(O_{2}\) gas (relative molar mass 32 ) at a temperature \(\mathrm{T}\). The pressure of the gas is \(\mathrm{P}\). An identical vessel containing 1 mole of He gas (relative molar mass 4 ) at a temperature \(2 \mathrm{~T}\) has pressure of \(.\) (A) \(8 \mathrm{P}\) (B) \((\mathrm{p} / 8)\) (C) \(2 \mathrm{P}\) (D) \(\mathrm{P}\)
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