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Question
Consider a mixture of 2 mol of helium and 4 mol of oxygen. Compute the speed of sound in this gas mixture at 300 K.
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Solution
Number of molecules of helium = 2
Number of molecules of oxygen = 4
When helium and oxygen are mixed, hence the molecular weight of the mixture of gases is given by
`"M"_"mix" = [("n"_1"M"_1 + "n"_2"M"_2)/("n"_1 + "n"_2)]`
`= [(2 xx 4 + 4 xx 32)/(2 + 4)]`kg/mol
`= (8 + 128)/6`
`= 136/6` kg/mol
= 22.6 × 10-3 kg/mol
In addition, helium is monoatomic,
`"C"_("v"_2) = (2"R")/2`
Oxygen is diatomic `"C"_("v"_1) = (5"R")/2`
∴ For the mixture `("C"_"v")_"mix" = ("n"_1"C"_("v"_1) + "n"_2"C"_("v"_2))/("n"_1 + "n"_2)`
`("C"_"v")_"mix" = (2 xx 3/2 "R" + 4 xx 5/2 "R")/(2 + 4) = (13 "R")/6`
From Meyor's relation
`("C"_"p")_"mix" = ("C"_"v")_"mix" + "R"`
`("C"_"p")_"mix" = (13"R")/6 + "R" = (13"R" + "5R")/6 = (19"R")/6`
Ratio of specific heat capacitors of a mixture of gases is
`lambda_"mix" = "C"_"p"/"C"_"v" = ((19"R")/6)/((13"R")/6) = 19/13`
According to Laplace, the speed of sound in a gas is
v = `sqrt((lambda "RT")/("M"))`
v = `sqrt(19/13 xx (8.31 xx 300)/(22.6 xx 10^-3))`
`= sqrt(28420.2/17.68 xx 10^4)`
= 4.009 × 102 m/s
= 400.9 m/s
∴ The speed of sound = 400.9 m/s
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