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Chapter 9: Problem 1292

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 = \(\dfrac{\mathrm{v}_{\mathrm{rms2}} - \mathrm{v}_{\mathrm{rms1}}}{\mathrm{v}_{\mathrm{rms1}}}\) × 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 = \(\dfrac{\sqrt{T_2} - \sqrt{T_1}}{\sqrt{T_1}}\) × 100 % Substituting the values of \(T_1\) and \(T_2\) obtained in Step 1: Percentage increase = \(\dfrac{\sqrt{400.15} - \sqrt{300.15}}{\sqrt{300.15}}\) × 100 %
04

Evaluate the percentage increase

Now, we can compute the percentage increase: Percentage increase ≈ \(\dfrac{20.004 - 17.324}{17.324}\) × 100 % Percentage increase ≈ 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 product of the pressure and volume of an ideal gas \(1 \mathrm{~s}\) (A) A constant (B) Directly proportional to its temperature. (C) Inversely proportional to its temperature. (D) Approx. equal to the universal gas constant.

The relation between the gas pressure \(\mathrm{P}\) and average kinetic energy per unit volume \(E\) is (A) \(\mathrm{P}=(2 / 3) \mathrm{E}\) (B) \(P=(3 / 2) E\) (C) \(\mathrm{P}=\mathrm{E}\) (D) \(\mathrm{P}=(\mathrm{E} / 2)\)

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 volume of a gas at \(20 \mathrm{C}\) is \(200 \mathrm{ml}\). If the temperature is reduced to \(-20^{\circ} \mathrm{C}\) at constant pressure, its volume will be. (A) \(172.6 \mathrm{~m} 1\) (B) \(17.26 \mathrm{ml}\) (C) \(19.27 \mathrm{ml}\) (D) \(192.7 \mathrm{ml}\)

What is the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2 atm and temperature $17^{\circ} \mathrm{C} ?$ Take the radius of nitrogen molecule to be 1A. Molecular mass of nitrogen \(=28\), $\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{JK}^{-1}, 1 \mathrm{~atm}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$ (A) \(2.2 \times 10^{-7} \mathrm{~m}, 2.58 \times 10^{9}\) (B) \(1.1 \times 10^{-7} \mathrm{~m}, 4.58 \times 10^{8}\) (C) \(1.1 \times 10^{-7} \mathrm{~m}, 4.58 \times 10^{9}\) (D) \(2.2 \times 10^{-7} \mathrm{~m}, 3.58 \times 10^{9}\)
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