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Step-by-Step Solution
Step 1: Recall the formula for the speed of sound in an ideal gas
The speed of sound in an ideal gas is given by
$ v = \sqrt{\frac{\gamma R T}{M}}, $
where:
$\gamma$ is the ratio of specific heats (also known as the adiabatic index).
$R$ is the universal gas constant.
$T$ is the absolute temperature.
$M$ is the molar mass of the gas.
Step 2: Observe that temperature is the same for both gases
Since the temperature $T$ is the same for both oxygen $(O_2)$ and helium $(He)$, the speed of sound in each gas primarily depends on the ratio $\frac{\gamma}{M}$.
Thus,
$ v \propto \sqrt{\frac{\gamma}{M}}. $
Step 3: Write the ratio of speeds in the two gases
Let $v_{O_2}$ be the speed of sound in oxygen and $v_{He}$ be the speed of sound in helium. Then:
$ \frac{v_{O_2}}{v_{He}}
= \sqrt{ \frac{\gamma_{O_2}/M_{O_2}}{ \gamma_{He}/M_{He}} }
= \sqrt{ \frac{\gamma_{O_2}}{\gamma_{He}} \times \frac{M_{He}}{M_{O_2}} }. $
Step 4: Substitute numerical values
For diatomic oxygen $(O_2)$:
$ \gamma_{O_2} = 1.4, \quad M_{O_2} = 32 \, ( \text{g/mol}, \text{if using SI, consider 0.032 kg/mol} ).$
For helium $(He)$:
$ \gamma_{He} = 1.67, \quad M_{He} = 4 \, ( \text{g/mol}, \text{or } 0.004 \text{ kg/mol} ).$
Thus,
$
\frac{v_{O_2}}{v_{He}}
= \sqrt{ \frac{1.4}{1.67} \times \frac{4}{32} }
= \sqrt{ \frac{1.4}{1.67} \times \frac{1}{8} }
= 0.3237 \text{ (approximately)}.
$
Step 5: Solve for $v_{He}$
We know $v_{O_2} = 460 \, \text{m/s}$. From the above ratio:
$
\frac{v_{O_2}}{v_{He}} = 0.3237
\quad \implies \quad
v_{He} = \frac{v_{O_2}}{0.3237} = \frac{460}{0.3237} \approx 1421 \, \text{m/s}.
$
Answer
$1421\, \text{m/s}$.
