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For fire safety considerations, there is a proposal to replace nitrogen with helium in air as...

For fire safety considerations, there is a proposal to replace nitrogen with helium in air as the atmosphere of EVA operation in manned space vehicles. Compute the speed of sound for an atmosphere with 25% O2 and 75% He (volume or molar percentages) at a temperature of 15 C and 80 kPa pressure. Compare your answer with that of standard air in the same environmental conditions. (Hint: you need to determine both the molecular weight of the mixture and the ratio of specific heats--use molar weighted averages)

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Solution.

For an ideal gas the sound speed can be written as

c = \sqrt{\frac{k P}{\rho}}

where ρ is the mass density. The constant k is present because the expansion and compression of the gas is supposed to be adiabatic, and kk is the adiabatic index. For a gas mixture with n moles of the i-th gas we have

k= \frac{\sum_i n_i \frac{k_i}{k_i+1}}{\sum_i n_i \frac{1}{k_i+1}}

For an ideal monoatomic gas, the adiabatic index is 5/3 k(He) = 5/3;

For an ideal monoatomic gas, the adiabatic index is k(O2) = 7/5;

k= \frac{0.25\times \frac{7/5}{1+7/5} + 0.75\times \frac{5/3}{1+5/3}}{0.25\times \frac{1}{1+7/5} + 0.75\times \frac{1}{1+5/3}} = 1.595;

\rho = 0.25\times \frac{0.032}{V_{O2}} + 0.75\times \frac{0.004}{V_{He}};

\frac{V_{O2}}{V_{He}} = \frac{0.25}{0.75};

{V_{O2}} + {V_{He}} = \frac{RT}{P};

P = 80000 Pa
Converting temperature to SI units.
T = 288.15 K
n = 1
R = 8.31 J/(mol*K)
Computing volume using the selected equation.
V = 0.03 m3;

V(O2) = 0.0075m3; V(He) = 0.0225m3;

\rho = 0.25\times \frac{0.032}{0.0075} + 0.75\times \frac{0.004}{0.0225} = 1.2 \ \frac{kg}{m^{3}};

The speed of sound for an atmosphere with 25% O2 and 75% He (volume or molar percentages) at a temperature of 15 C and 80 kPa pressure is

c = \sqrt{\frac{1.595\times 80000}{1.2}} = 326.1 \ \frac{m}{s};

The speed of sound for an atmosphere with the standard air in the same environmental conditions is

c_{air} = \sqrt{\frac{kRT}{M}} = \sqrt{\frac{1.4\times 8.314\times 288.15}{0.02895}} = 340.4 \ \frac{m}{s}.

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