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Chapter 6: Problem 74

The \(u_{\mathrm{rms}}\) of \(\mathrm{H}_{2}\) molecules at \(273 \mathrm{K}\) is \(1.84 \times 10^{3} \mathrm{m} / \mathrm{s}\) At what temperature is \(u_{\mathrm{rms}}\) for \(\mathrm{H}_{2}\) twice this value?

Short Answer

Expert verified
The temperature at which the root mean square speed of a \(H_{2}\) molecule is twice the given value is \(1092 K\).

Step by step solution

01

Use the rms speed equation to find the molar mass

First, it's necessary to determine the molar mass of the \(H_{2}\) molecule using the given root mean square speed at \(273 K\). This can be done by rearranging the rms speed formula solving for m: \(m = \frac{3kT}{u_{rms}^{2}}\) where \(u_{rms} = 1.84 \times 10^{3} m/s\), T = 273 K, and \(k = 1.38 \times 10^{-23} J/K\).
02

Calculate the molar mass

Substituting the values into the equation gives \(m = \frac{3 \times 1.38 \times 10^{-23} J/K \times 273 K}{(1.84 \times 10^{3} m/s)^{2}} = 2.82 \times 10^{-27} kg\)
03

Use the rms speed equation to find the temperature when the speed is doubled

The question asks for the temperature when the root mean square speed is twice the given value, so \(u_{rms} = 2 \times 1.84 \times 10^{3} m/s\). Substitute these values into the rms speed formula while solving for T: \(T = \frac{m \times (u_{rms})^{2}}{3k}\)
04

Calculate the temperature

Substitute the values into the equation from Step 3: \(T = \frac{2.82 \times 10^{-27} kg \times (2 \times 1.84 \times 10^{3} m/s)^{2}}{3 \times 1.38 \times 10^{-23} J/K} = 1092 K\)

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