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Chapter 9: Problem 1304

The temperature of an ideal gas is increased from \(27^{\circ} \mathrm{C}\) to \(927^{\circ} \mathrm{C}\). The root mean square speed of its molecules becomes (A) Four times (B) One-fourth (C) Half (D) Twice

Short Answer

Expert verified
The root-mean-square speed of the gas molecules becomes twice when the temperature is increased from 27°C to 927°C. The correct answer is (D) Twice.

Step by step solution

01

Convert temperatures to Kelvin

We will first convert the given temperatures from Celsius to Kelvin as temperature in Kelvin is used in rms speed formula. We convert by adding 273.15 to the Celsius temperature: \( T_1 = 27°C + 273.15 = 300.15 K \) \( T_2 = 927°C + 273.15 = 1200.15 K \)
02

Recall the formula for rms speed of an ideal gas

The formula for the root-mean-square speed (v_rms) of an ideal gas is: \( v_{rms} = \sqrt{\frac{3RT}{M}} \) where: - v_rms: root-mean-square speed of the gas molecules - R: universal gas constant (8.314 J/mol·K) - T: temperature of the gas in Kelvin - M: molar mass of the gas molecules (in kg/mol)
03

Compare initial and final rms speeds

Here we are only concerned about the ratio of final rms speed to the initial rms speed. So, let's find the ratio: \( \frac{v_{rms2}}{v_{rms1}} = \frac{\sqrt{\frac{3RT_2}{M}}}{\sqrt{\frac{3RT_1}{M}}} \) As both numerator and denominator share the same terms (3, R, and M), we can simplify above expression: \( \frac{v_{rms2}}{v_{rms1}} = \frac{\sqrt{T_2}}{\sqrt{T_1}} \)
04

Calculate the ratio of final and initial rms speeds

Substitute the initial and final temperatures (T1 and T2) that we calculated in Step 1: \( \frac{v_{rms2}}{v_{rms1}} = \frac{\sqrt{1200.15}}{\sqrt{300.15}} \) Now, calculate the ratio: \( \frac{v_{rms2}}{v_{rms1}} \approx 2 \)
05

Find the answer

Since the ratio of the final rms speed to the initial rms speed is approximately 2, it means the root-mean-square speed of the gas molecules becomes twice when the temperature is increased from 27°C to 927°C. So the correct answer is: (D) Twice

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Most popular questions from this chapter

The molecules of a given mass of a gas have a rms velocity of $200 \mathrm{~m} / \mathrm{s}\( at \)27^{\circ} \mathrm{C}\( and \)1.0 \times 10^{5} \mathrm{Nm}^{-2}\( pressure when the temperature is \)127^{\circ} \mathrm{C}$ and pressure is \(0.5 \times 10^{5} \mathrm{Nm}^{-2}\), the rms velocity in \(\mathrm{m} / \mathrm{s}\) will be (A) \(100 \sqrt{2}\) (B) \([(100 \sqrt{2}) / 3]\) (C) \([(400) / \sqrt{3}]\) (D) None of these

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

The temperature of a gas at pressure \(\mathrm{P}\) and volume \(\mathrm{V}\) is \(27^{\circ} \mathrm{C}\) Keeping its volume constant if its temperature is raised to \(927^{\circ} \mathrm{C}\), then its pressure will be (A) \(3 \mathrm{P}\) (B) \(2 \mathrm{P}\) (C) \(4 \mathrm{P}\) (D) \(6 \mathrm{P}\)

To what temperature should the hydrogen at room temperature $\left(27^{\circ} \mathrm{C}\right)$ be heated at constant pressure so that the rms velocity of its molecule becomes double of its previous value (A) \(927^{\circ} \mathrm{C}\) (B) \(600^{\circ} \mathrm{C}\) (C) \(108^{\circ} \mathrm{C}\) (D) \(1200^{\circ} \mathrm{C}\)

Suppose ideal gas equation follows \(V P^{3}=\) constant, Initial temperature and volume of the gas are \(\mathrm{T}\) and \(\mathrm{V}\) respectively. If gas expand to \(27 \mathrm{~V}\), then temperature will become (A) \(9 \mathrm{~T}\) (B) \(27 \mathrm{~T}\) (C) (T/9) (D) \(\mathrm{T}\)
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