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Chapter 9: Problem 1280

A sample of gas is at \(0^{\circ} \mathrm{C}\). To what temperature it must be raised in order to double the rms speed of molecule. (A) \(270^{\circ} \mathrm{C}\) (B) \(819^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(1090^{\circ} \mathrm{C}\)

Short Answer

Expert verified
To double the rms speed of a gas initially at \(0^{\circ}\mathrm{C}\), it must be raised to \(819^{\circ}\mathrm{C}\).

Step by step solution

01

Isolate the temperature terms

In the proportion relating the initial and final rms speeds, we can eliminate the gas constant (R) and the molar mass (M) of the gas by dividing the entire equation by the initial speed: \(\frac{v_{rms}'}{v_{rms}} = \frac{\sqrt{3RT'}}{\sqrt{3RT}}\) Now, square both sides of the equation to get rid of the square roots: \(\left(\frac{v_{rms}'}{v_{rms}}\right)^2 = \frac{3RT'}{3RT}\) We are given that the final rms speed is twice the initial speed, so we can substitute 2 for \(\frac{v_{rms}'}{v_{rms}}\), giving us: \((2)^2 = \frac{T'}{T}\)
02

Solve for the final temperature T'

Now that we have an equation that relates the initial and final temperatures, we can solve for the final temperature: \((2)^2 = \frac{T'}{T} \Rightarrow 4T = T'\) We are given that the initial temperature is \(0^{\circ}\mathrm{C}\), which is equivalent to 273 K. Substitute this value into the equation to find the final temperature: \(4(273\,\mathrm{K}) = T'\) \(T' = 1092\,\mathrm{K}\)
03

Convert the final temperature to Celsius

To convert the final temperature from Kelvin to Celsius, subtract 273 from the value of T': \(1092\,\mathrm{K} - 273 = 819^{\circ}\mathrm{C}\) So the answer is: (B) \(819^{\circ}\mathrm{C}\)

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Most popular questions from this chapter

Air is pumped into an automobile tube upto a pressure of \(200 \mathrm{kPa}\) in the morning when the air temperature is \(22^{\circ} \mathrm{C}\). During the day, temperature rises to \(42^{\circ} \mathrm{C}\) and the tube expands by $2 \%$ The pressure of the air in the tube at this temperature will be approximately. (A) \(209 \mathrm{kPa}\) (B) \(206 \mathrm{kPa}\) (C) \(200 \mathrm{kPa}\) (D) \(212 \mathrm{kPa}\)

What is the mass of 2 liters of nitrogen at \(22.4\) atmospheric pressure and \(273 \mathrm{~K} .\left(\mathrm{R}=8.314 \mathrm{Jmol} \mathrm{k}^{-1}\right)\) (A) \(14 \times 22.4 \mathrm{~g}\) (B) \(56 \mathrm{~g}\) (C) \(28 \mathrm{~g}\) (D) none of these

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

When the temperature of a gas is raised from \(27^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\), the percentage increase in the rms velocity of the molecules will be (A) \(15 \%\) (B) \(17.5 \%\) (C) \(10 \%\) (D) \(20 \%\)
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