The ratio of the speed of sound in \(\mathrm{H_2}\) (molecular weight \(2~\mathrm{kg/kmol}\)) to that in \(\mathrm{N_2}\) (molecular weight \(28~\mathrm{kg/kmol}\)) at \(T=300~\mathrm{K}\) and \(p=2~\mathrm{bar}\) is \underline{\hspace{1cm}}. \;(round off to two decimal places)
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Solution and Explanation
Step 1: Relation for sound speed in an ideal gas.
\[
a \;=\; \sqrt{\gamma R T} \;=\; \sqrt{\gamma \,\frac{R_u}{M}\,T}
\]
where \(\gamma\) is ratio of specific heats, \(R_u=8.314~\mathrm{kJ/(kmol\cdot K)}\), and \(M\) is the molecular weight.
\(\Rightarrow a \propto \sqrt{\displaystyle \frac{\gamma}{M}}\) at fixed \(T\).
Step 2: Apply to \(\mathrm{H_2}\) and \(\mathrm{N_2}\).
At \(300~\mathrm{K}\) both gases are well–approximated as diatomic with \(\gamma \approx 1.4\) (small differences in \(\gamma\) change the answer only in the 3rd decimal). Hence
\[
\frac{a_{\mathrm{H_2}}}{a_{\mathrm{N_2}}}
=\sqrt{\frac{\gamma_{\mathrm{H_2}}/M_{\mathrm{H_2}}}{\gamma_{\mathrm{N_2}}/M_{\mathrm{N_2}}}}
\approx \sqrt{\frac{1.4/2}{1.4/28}}
=\sqrt{\frac{28}{2}}
=\sqrt{14}
=3.741657\ldots
\]
Step 3: Pressure irrelevance check.
Since \(a\) depends only on \(T,\gamma,M\) for an ideal gas, the given pressure (\(2~\mathrm{bar}\)) does not affect the ratio.
Final Answer:
\[
\boxed{3.74}
\]
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