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Chapter 9: Problem 1290

When temperature of an ideal gas is increased from \(27^{\circ} \mathrm{C}\) to \(227^{\circ} \mathrm{C}\), its rms speed changed from \(400 \mathrm{~ms}^{-1}\) to \(\mathrm{V}_{\mathrm{s}}\). The \(\mathrm{V}_{\mathrm{s}}\) is (A) \(516 \mathrm{~ms}^{-1}\) (B) \(746 \mathrm{~ms}^{-1}\) (C) \(310 \mathrm{~ms}^{-1}\) (D) \(450 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The correct answer is (A) \(516 \mathrm{~ms}^{-1}\). This is found by using the equation \(\frac{v_{rms2}}{v_{rms1}} = \sqrt{\frac{T_2}{T_1}}\), converting temperatures to Kelvins, and substituting the given values.

Step by step solution

01

Understand the rms speed and temperature relationship

For an ideal gas, the rms speed is related to its temperature by the equation: \(v_{rms} = \sqrt{\dfrac{3kT}{m}}\) where \(v_{rms}\) is the rms speed, \(k\) is the Boltzmann constant, \(T\) is the temperature in Kelvins, and \(m\) is the mass of a single gas molecule. The ratio of the rms speeds at two different temperatures can be found by dividing the equation at the final temperature by the equation at the initial temperature: \(\dfrac{v_{rms2}}{v_{rms1}} = \sqrt{\dfrac{3kT_2/m}{3kT_1/m}}\) The mass of a single gas molecule and the Boltzmann constant are the same for both temperatures, so the constants can cancel out, leaving: \(\dfrac{v_{rms2}}{v_{rms1}} = \sqrt{\dfrac{T_2}{T_1}}\)
02

Convert the temperatures to Kelvins

To solve the equation, first convert the given temperatures from Celsius to Kelvins: Initial temperature: \(T_1 = 27°C + 273.15 = 300.15 K\) Final temperature: \(T_2 = 227°C + 273.15 = 500.15 K\)
03

Substitute the given values into the equation

Now, substitute the given values and the converted temperatures into the equation: \(\dfrac{v_{rms2}}{400 m/s} = \sqrt{\dfrac{500.15 K}{300.15 K}}\)
04

Solve for the final rms speed (Vs)

Rearrange the equation to find the final rms speed \(v_{rms2}\): \(v_{rms2} = 400 m/s \times \sqrt{\dfrac{500.15 K}{300.15 K}}\) Calculate the value of \(v_{rms2}\): \(v_{rms2} \approx 516.14 m/s\)
05

Match the answer to the given options

The calculated value of the final rms speed (Vs) matches the option (A) \(516 \mathrm{~ms}^{-1}\). So, the correct answer is (A) \(516 \mathrm{~ms}^{-1}\).

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Most popular questions from this chapter

A perfect gas at \(27^{\circ} \mathrm{C}\) is heated at constant pressure to \(327^{\circ} \mathrm{C}\). If original volume of gas at \(27^{\circ} \mathrm{C}\) is \(\mathrm{V}\) then volume at \(327^{\circ} \mathrm{C}\) is (A) \(2 \mathrm{~V}\) (B) \(\mathrm{V}\) (C) (V/2) (D) \(3 \mathrm{~V}\)

The average kinetic energy of a gas molecule at \(27^{\circ} \mathrm{C}\) is \(6.21 \times 10^{-21} \mathrm{~J}\). Its average kinetic energy at $227^{\circ} \mathrm{C}$ will be (A) \(5.22 \times 10^{-21} \mathrm{~J}\) (B) \(11.35 \times 10^{-21} \mathrm{~J}\) (C) \(52.2 \times 10^{-21} \mathrm{~J}\) (D) \(12.42 \times 10^{-21} \mathrm{~J}\)

The root mean square speed of hydrogen molecules at \(300 \mathrm{~K}\) is $1930 \mathrm{~m} / \mathrm{s}$. Then the root mean square speed of Oxygen molecules at \(900 \mathrm{~K}\) will be (A) \(836 \mathrm{~m} / \mathrm{s}\) (B) \(643 \mathrm{~m} / \mathrm{s}\) (C) \(1930 \sqrt{3} \mathrm{~m} / \mathrm{s}\) (D) \([(1930) / \sqrt{3}] \mathrm{m} / \mathrm{s}\)

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) \(192.7 \mathrm{ml}\)

The relation between two specific heats of a gas is (A) $\mathrm{C}_{\mathrm{V}}-\mathrm{C}_{\mathrm{P}}=(\mathrm{R} / \mathrm{J})$ (B) \(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{J}\) (C) $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=(\mathrm{R} / \mathrm{J})$ (D) \(\mathrm{C}_{\mathrm{V}}-\mathrm{C}_{\mathrm{P}}=\mathrm{J}\)
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