The ratio of the densities of oxygen and nitrogen is `16 : 14` . Calculate the temperature when the speed of sound in nitrogen gas at `17^(@) C` is equal to the speed of sound in oxygen gas .

Explanation

From equation , we have `v = sqrt((gamma P)/(rho))`
But ` rho = (M)/(V)`
Therefore , `v = sqrt((gamma PV)/(M))`
Using equation , `v = sqrt((gamma RT)/(M))`
Where , R is the universal gas constant and M is the molecular mass of the gas . The speed of sound in nitrogen gas at `17^(@) C` is
`v_(N) = sqrt((gamma R (273 K + 17K))/(M_(N))) = sqrt((gamma R(290K))/(M_(N))) " " ... (1)`
Similarly , the speed of sound in oxygen gas at t in K is
`v_(O) = sqrt((gamma R (273 + t))/(M_(O))) " " ... (2)`
Given that the value of `gamma` is same for both the gases , the two speeds must be equal . Hence , equating equation (1) and (2) , we get
`v_(O) = v_(N)`
`sqrt((gamma R (273 + t))/(M_(O))) = sqrt((gamma R (290))/(M_(N)))`
Squaring on both sides and cancelling `gamma R` term and rearranging , we get
`(M_(O))/(M_(N)) = (273 + t)/(290) " " .... (3)`
Since the densities of oxygen and nitrogen is `16 : 14` ,
`(rho_(O))/(rho_(N)) = (16)/(14) " " ... (4)`
`(rho_(O))/(rho_(N)) = ((M_(O))/(V))/((M_(N))/(V)) = (M_(O))/(M_(N)) implies (M_(O))/(M_(N)) = (16)/(14)`
Substituting equation (5) in equation (3) , we get
`(273 + t)/(290) = (16)/(14) implies 3822 + 14 t = 4640 implies t = 58.4 K`