Question

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)

Hint:

At the same temperature, for gases with similar \(\gamma\), \(a \propto 1/\sqrt{M}\). Lighter gas \(\Rightarrow\) higher sound speed.

Answer 1

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:
\[ \boxed{3.74} \]