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Chapter 9: Problem 1293

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) \(37 \%\)

Short Answer

Expert verified
The percentage increase in the root mean square velocity (\(\mathrm{v}_{\mathrm{rms}}\)) is approximately 15.5%.

Step by step solution

01

Convert temperatures to Kelvin

Since the given temperatures are in Celsius, we need to convert them to Kelvin by adding 273.15 to each of them. For the initial temperature, \(T_1 = 27 ^{\circ} \mathrm{C}\) + 273.15 = 300.15 K For the final temperature, \(T_2 = 127 ^{\circ} \mathrm{C}\) + 273.15 = 400.15 K
02

Find the initial and final root mean square velocities

The formula for \(\mathrm{v}_{\mathrm{rms}}\) is \(\sqrt{\dfrac{3 RT}{M}}\). Since we are considering the same gas, we can focus on the temperature term. The initial root mean square velocity, \(\mathrm{v}_{\mathrm{rms1}}\) is proportional to \(\sqrt{T_1}\) The final root mean square velocity, \(\mathrm{v}_{\mathrm{rms2}}\) is proportional to \(\sqrt{T_2}\)
03

Compute the percentage increase in root mean square velocity

To calculate the percentage increase in \(\mathrm{v}_{\mathrm{rms}}\), we will use the following formula: Percentage increase = \(\dfrac{\mathrm{v}_{\mathrm{rms2}} - \mathrm{v}_{\mathrm{rms1}}}{\mathrm{v}_{\mathrm{rms1}}}\) 脳 100 % Since \(\mathrm{v}_{\mathrm{rms1}}\) and \(\mathrm{v}_{\mathrm{rms2}}\) are proportional to \(\sqrt{T_1}\) and \(\sqrt{T_2}\) respectively, we can write: Percentage increase = \(\dfrac{\sqrt{T_2} - \sqrt{T_1}}{\sqrt{T_1}}\) 脳 100 % Substituting the values of \(T_1\) and \(T_2\) obtained in Step 1: Percentage increase = \(\dfrac{\sqrt{400.15} - \sqrt{300.15}}{\sqrt{300.15}}\) 脳 100 %
04

Evaluate the percentage increase

Now, we can compute the percentage increase: Percentage increase 鈮� \(\dfrac{20.004 - 17.324}{17.324}\) 脳 100 % Percentage increase 鈮� 15.5 %
05

Choose the correct answer

We find that the percentage increase in \(\mathrm{v}_{\mathrm{rms}}\) is about 15.5 %. Thus, the correct answer is (C) 15.5 %.

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Most popular questions from this chapter

The degrees of freedom for triatomic gas \(1 \mathrm{~s}\) (At room temperature) (A) 8 (B) 6 (C) 4 (D) 2

If three molecules have velocities \(0.5,1\) and 2 the ratio of rms speed and average speed is (The velocities are in \(\mathrm{km} / \mathrm{s}\) ) (A) \(0.134\) (B) \(1.34\) (C) \(1.134\) (D) \(13.4\)

At what temperature pressure remaining constant will the \(\mathrm{rms}\) speed of a gas molecules increase by \(10 \%\) of the \(\mathrm{rms}\) speed at NTP? (A) \(57.3 \mathrm{~K}\) (B) \(57.3^{\circ} \mathrm{C}\) (C) \(557.3 \mathrm{~K}\) (D) \(-57.3^{\circ} \mathrm{C}\)

The pressure and temperature of two different gases \(P\) and T having the volumes \(\mathrm{V}\) for each. They are mixed keeping the same volume and temperature, the pressure of the mixture will be, (A) \(\mathrm{P}\) (B) \((\mathrm{P} / 2)\) (C) \(4 \mathrm{P}\) (D) \(2 \mathrm{P}\)

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) \(1.73 \sqrt{\left[M_{0} /(R T)\right]}\)
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