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Under standard conditions, the density of a gas is ( 1400 1089 ~kg - m ⁻³ ) and the speed of sound propagation in it is (330 ~ms ⁻¹ ), then the number of degrees of freedom of the gas molecules is

Options

  1. A2
  2. B7
  3. C5
  4. D3

Correct answer

C. 5

Step-by-step solution

Given, density of gas, ( = 1400 1089 ~kg / m ^3 ) speed of sound, (v=330 ~m / s ) and under standard condition, Pressure of gas, (p=1 10^5 ~N / m ^2 ) If ( ) be the ratio of (C_p ) and (C_V ) of a gas, then the speed of sound in gas is given by ( aligned & v= P or P =v^2 & = v^2 p = 330 330 10^5 1400 1089 & =1.4 & So, = C_p C_V =1.4 aligned [ array l =1+ 2 f Since, for diatomic gas, the volume of array ] aligned & is 1.4. & Hence, the degree of freedom for aligned ) So, ( = C_p C_V =1.4 ) Since, for diatomic gas, t

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