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Chapter 9: Problem 1296

The rms speed of the molecules of a gas at a pressure \(10^{5} \mathrm{~Pa}\) and temperature \(0^{\circ} \mathrm{C}\) is \(0.5 \mathrm{~km} / \mathrm{s}\). If the pressure is kept constant but temperature is raised to $819^{\circ} \mathrm{C}$, the rms speed becomes (A) \(1.5 \mathrm{kms}^{-1}\) (B) \(2 \mathrm{kms}^{-1}\) (C) \(1 \mathrm{kms}^{-1}\) (D) \(5 \mathrm{kms}^{-1}\)

Short Answer

Expert verified
The final rms speed of the gas molecules at the given conditions is 1 km/s. Therefore, the correct answer is (C) \(1 \mathrm{kms}^{-1}\).

Step by step solution

01

Determine the formula relating rms speed, temperature, and pressure

The relationship between the root-mean-square speed (\(u\)), the gas constant(\( R\)), the molar mass (\(M\)), and the temperature (\(T\)) is given by: \(u = \sqrt{\frac{3RT}{M}}\)
02

Convert temperatures to Kelvin

To work with temperature in the equation, we need to convert it to Kelvin by adding 273.15 to the Celcius values: \(T_1 = 0^{\circ} \mathrm{C} + 273.15 = 273.15\mathrm{K}\) \(T_2 = 819^{\circ} \mathrm{C} + 273.15 = 1092.15\mathrm{K}\)
03

Express the given information as equations

We have the initial rms speed (\(u_1\)) and the initial temperature (\(T_1\)). Similarly, we need to find the final rms speed (\(u_2\)) at the final temperature (\(T_2\)). The equations are: \(u_1 = \sqrt{\frac{3R*T_1}{M}}\) \(u_2 = \sqrt{\frac{3R*T_2}{M}}\)
04

Divide the second equation by the first equation

To eliminate the constants, we divide the second equation by the first equation: \(\frac{u_2}{u_1} = \sqrt{\frac{T_2}{T_1}}\)
05

Substitute the given values and solve for \(u_2\)

We know that \(u_1\) = 0.5 km/s = 500 m/s, \(T_1\) = 273.15 K, and \(T_2\) = 1092.15 K. We can now substitute these values into the equation and solve for \(u_2\): \(\frac{u_2}{500\,m/s} = \sqrt{\frac{1092.15\,K}{273.15\,K}}\) \(u_2 = 500\,m/s * \sqrt{\frac{1092.15\,K}{273.15\,K}}\)
06

Calculate the final rms speed

Perform the calculation and convert the result back to km/s: \(u_2 = 500\,m/s * \sqrt{\frac{1092.15\,K}{273.15\,K}} = 1000 \,\mathrm{m/s} = 1\, \mathrm{km/s}\) The final rms speed of the gas molecules at the given conditions is 1 km/s. Therefore, the correct answer is (C) \(1 \mathrm{kms}^{-1}\).

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