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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. - Physics

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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.

Numerical
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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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Chapter 11: Waves - Evaluation [Page 278]

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Samacheer Kalvi Physics - Volume 1 and 2 [English] Class 11 TN Board
Chapter 11 Waves
Evaluation | Q IV. 2. | Page 278

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