Question

Calculate the ratio of the speed of sound in neon to that in water vapour at any given temperature.
[molecular weight of neon $=2.02\times {10}^{-2}\text{kg/mol}$ and of water vapour $=1.8\times {10}^{-2}\text{kg/mol}$. Given that water vapour is triatomic]

• A. $1.023$
• B. $1.5$
• C. $1.055$
• D. $1.8$

Answer 1

The correct option is C $1.055$

Given that,
Molecular weight of the neon, $M_{Ne}=2.02\times {10}^{-2}\text{kg/mol}$
Molecular weight of the water vapour, $M_W=1.8\times {10}^{-2}\text{kg/mol}$

The velocity of sound in a medium is given by
$v=\sqrt{\frac{E}{\rho }}=\sqrt{\frac{\gamma P}{\rho }}=\sqrt{\frac{\gamma RT}{M}}$

Let the velocity of sound in neon be $v_{Ne}$ and water be $v_W$.
In terms of temperature and molecular weight,
$\frac{v_{Ne}}{v_W}=\sqrt{\frac{{\gamma }_{Ne}}{M_{Ne}}\times \frac{M_W}{{\gamma }_W}}$ [as $T_{Ne}=T_W$]
Now, as neon is monoatomic i.e $(\gamma =5/3)$ while water vapour is triatomic i.e $(\gamma =4/3)$, so
$\frac{v_{Ne}}{v_W}=\sqrt{\frac{\left(5/3\right)\times 1.8\times {10}^{-2}}{\left(4/3\right)\times 2.02\times {10}^{-2}}}=\sqrt{\frac{5}{4}\times \frac{1.8}{2.02}}=1.055$