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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
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If under standard conditions a gas of density $1.2 kg/m^3$ has sound propagating in it with velocity of $340 m/s$, what is its atomicity/

$\begin{array}{1 1} diatomic \\ monoatomic \\ triatomic \\ none\;of\;these \end{array}$

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  • For a gas, the speed of sound $v_{\text{sound}} = \large \sqrt (\gamma \large \frac{p}{\rho})$, where $\gamma$ is the adiabatic index, $p$ is the pressure and $\rho$ is the density.
  • $\gamma$ is the adiabatic index also known as the isentropic expansion factor. For diatomic molecules, this value is 1.4.
Given, density $\rho = 1.2 kg/m^3$ and $v_{\text{sound}} = 340 m/s$. Pressure $= 1.01 \times 10^5 P_a$
For a gas, the speed of sound $v_{\text{sound}} = \large \sqrt (\gamma \large \frac{p}{\rho})$, where $\gamma$ is the adiabatic index, $p$ is the pressure and $\rho$ is the density.
If $v_{\text{sound}} = \large \sqrt (\gamma \large \frac{p}{\rho}) \rightarrow \gamma = \large\frac{ (v_{\text{sound}})^2 \times \rho}{p}$
$\Rightarrow \gamma = \large\frac{340^2 \times 1.2}{1.01 \times 10^5} $$\approx 1.38 = 1.4$
$\gamma$ is the adiabatic index also known as the isentropic expansion factor. For diatomic molecules, this value is 1.4.
Answer: Since $\gamma = 1.4$, it is a diatomic molecule.
answered Jan 22, 2014 by balaji.thirumalai
 

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