91Ó°ÊÓ

The given data can be written as,

• Temperature is given as,\(T = 27^\circ \;{\rm{C}} = \left( {27 + 273} \right)\;{\rm{K}} = {\rm{300}}\;{\rm{K}}\).
• The molar mass of hydrogen is given as,\(M = 2.02\;{\rm{g/mol}} = 2.02 \times {10^{ - 3}}\;{\rm{kg/mol}}\) .

The expression for the velocity of the longitudinal wave traveling in a medium can be written as,

\(v = \sqrt {\frac{{\gamma RT}}{M}} \) (1)

Here\(\gamma \)is the adiabatic constant for the medium,\(R\)is the gas constant,\(T\)is the temperature, and \(M\)is the molar mass of the medium.

If we know these physical quantities, we can calculate the velocity of the wave easily.

The medium is hydrogen. The adiabatic constant for hydrogen is 1.41, and the gas constant is \(8.3145\;{\rm{J/mol}} \cdot {\rm{K}}\).

Substitute the values in equation 1, and we get the velocity of the wave in hydrogen as,

\(\begin{array}{c}{v_{{{\rm{H}}_{\rm{2}}}}} = \sqrt {\frac{{\left( {1.41} \right)\left( {8.3145\;{\rm{J/mol}} \cdot {\rm{K}}} \right)\left( {300\;{\rm{K}}} \right)}}{{2.02 \times {{10}^{ - 3}}\;{\rm{kg/mol}}}}} \\ = \sqrt {\frac{{\left( {1.41} \right)\left( {8.3145} \right)\left( {300} \right)}}{{2.02 \times {{10}^{ - 3}}}} \cdot \left( {\frac{{1\;{\rm{J/mol}} \cdot {\rm{K}}}}{{1\;{\rm{kg/mol}}}} \times 1\;{\rm{K}} \times \frac{{1\;{\rm{kg}} \cdot {{\rm{m}}^2} \cdot {{\rm{s}}^{ - 2}}}}{{1\;{\rm{J}}}}} \right)} \\ = \sqrt {\frac{{\left( {1.41} \right)\left( {8.3145} \right)\left( {300} \right)}}{{2.02 \times {{10}^{ - 3}}}} \cdot \left( {{\rm{1}}\;{{\rm{m}}^2} \cdot {{\rm{s}}^{ - 2}}} \right)} \\{v_{{{\rm{H}}_{\rm{2}}}}} = 1319\;{\rm{m/s}}\end{array}\)

Thus, the speed of longitudinal waves in hydrogen is \(1319\;{\rm{m/s}}\).