Question

At what temperature will the speed of sound in hydrogen be the same aas in oxygen at 100^(@) ? Molar Masses of oxygen and hydrogen are in the ratio 16: 1 .

Answer 1

Temperature will be -249.7° C or 23.2 K

• We know that,

        $V = \sqrt{\frac{\gamma *RT}{M} }$

• We want velocity to be same for both gases. γ and R are constants. To keep V constant if T increases, M should increase in same proportion so that they cancel each other out.

        ∴ T α M

            $\frac{T_h}{T_o} = \frac{M_h}{M_o}$

• Given temperature is in C. We convert the temperature to standard unit that is K.

        $\frac{T_h}{100+273} = \frac{2}{32}$

        On solving,

        ∴ T = 23.2 K

      = -249.7° C

Answer 2

Temperature of hydrogen molecule  = 23.2 k = -249.7°C

Given:

Oxygen at 100°

Molar Masses of oxygen and hydrogen are in the ratio 16: 1.

To find:

At what temperature will the speed of sound in hydrogen be the same aas in oxygen.

Formula used:

Velocity of sound in gas = $\sqrt{\frac{\gamma R T}{M} }$

Where $\gamma$ = Adiabatic constant.

            R = Gas constant.

            M = molecular mass.

            T = Temperature (kelvin)

Explanation:

Both hydrogen and oxygen are dia-atomic gas so that,

$\gamma$ = Adiabatic constant. is same for both the gas.

From above formula

Velocity of sound in gas = $\sqrt{\frac{\gamma R T}{M} }$

$\gamma$ , R are constant. and Speed of sound is same (Given)

So Temperature ∝ Molecular mass.

Temperature of oxygen = 100° c = 373k

$\frac{T_H}{T_O}$ = $\frac{M_H}{M_O}$

Temperature of hydrogen molecule =  $\frac{M_H}{M_O}$ × $T_o$

                                                             = $\frac{1}{16} \times 373$

Temperature of hydrogen molecule  = 23.2 k = -249.7°C

To learn more....

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