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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 average kinetic energy of hydrogen molecules at \(300 \mathrm{~K}\) is \(E\). At the same temperature the average kinetic energy of oxygen molecules will be (A) [E/(16)] (B) \(E\) (C) \(4 \mathrm{E}\) D) \([E /(4)]\)

Root mean square velocity of a molecule is \(v\) at pressure \(P\). If pressure is increased two times, then the rms velocity becomes (A) \(3 \mathrm{~V}\) (B) \(2 \mathrm{v}\) (C) \(0.5 \mathrm{~V}\) (D) \(\mathrm{v}\)

To what temperature should the hydrogen at \(327^{\circ} \mathrm{C}\) cooled at constant pressure, so that the root mean square velocity of its molecules become half of its previous value (A) \(-100^{\circ} \mathrm{C}\) (B) \(123^{\circ} \mathrm{C}\) (C) \(0^{\circ} \mathrm{C}\) (D) \(-123^{\circ} \mathrm{C}\)

The temperature of an ideal gas is increased from \(27^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\), then percentage increase in \(\mathrm{v}_{\mathrm{rms}}\) is (A) \(33 \%\) (B) \(11 \%\) (C) \(15.5 \%\) (D) \(37 \%\)

A cylinder of capacity 20 liters is filled with \(\mathrm{H}_{2}\) gas. The total average kinetic energy of translator motion of its molecules is $1.5 \times 10^{5} \mathrm{~J}$. The pressure of hydrogen in the cylinder is (A) \(4 \times 10^{6} \mathrm{Nm}^{-2}\) (B) \(3 \times 10^{6} \mathrm{Nm}^{-2}\) (C) \(5 \times 10^{6} \mathrm{Nm}^{-2}\) (D) \(2 \times 10^{6} \mathrm{Nm}^{-2}\)
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