Question

At what temperature will the speed of sound in a monoatomic gas be the same as in a diatomic gas at ${127}^{\circ }\text{C}$? Molar mass of diatomic and monoatomic gases are in the ratio $16:1$.

• A. ${21}^{\circ }\text{C}$
• B. $252\text{K}$
• C. $-{252}^{\circ }\text{C}$
• D. ${300}^{\circ }\text{C}$

Answer 1

The correct option is C $-{252}^{\circ }\text{C}$

We know that,
$\gamma$ of monoatomic gas $\left({\gamma }_m\right)=\frac{5}{3}$
$\gamma$ for diatomic gas $\left({\gamma }_d\right)=\frac{7}{5}$

From this, $\frac{{\gamma }_d}{{\gamma }_m}=\frac{21}{25}$

Let
$v_m=$ velocity of sound in monoatomic gas
$v_d=$ velocity of sound in diatomic gas

From the data given in the question,

Speed of sound in monotatomic gas =Speed of sound in diatomic gas.

(i.e) $v_m=v_d$

$\Rightarrow \sqrt{\frac{{\gamma }_{m}R{T}_m}{M_m}}=\sqrt{\frac{{\gamma }_{d}R{T}_d}{M_d}}$

$\Rightarrow \sqrt{\frac{T_m}{T_d}}=\sqrt{\frac{M_m}{M_d}}\times \sqrt{\frac{{\gamma }_d}{{\gamma }_m}}$
$\Rightarrow T_m=(\frac{M_m}{M_d}\times \frac{{\gamma }_d}{{\gamma }_m})T_d$

Given that, $\frac{M_d}{M_m}=\frac{16}{1}$

$T_d={127}^{\circ }\text{C}=127+273=400\text{K}$

So,
$T_m=[\frac{1}{16}\times \frac{21}{25}]\times 400=21\text{K}$

$\Rightarrow T_m=21\text{K or}-{252}^{\circ }\text{C}$

Thus, option (c) is the correct answer.