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

The ratio of the densities of oxygen and nitrogen is 16 : 14 . Calcula

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