Question

At what temperature does the average speed of an oxygen molecule equal that of an airplane moving at 590 mph ?

Answer 1

temperature does the average speed of an oxygen molecule equal that of an airplane moving at 590 mph = 89.24 K

Step-by-step explanation:

Average speed of oxygen molecule is given by

`v= \sqrt{(3RT)/(M) }`

`590*0.44704 = 263.75` m/s

R= 8.314 J/mol K = universal gas constant

M= molecular weight of oxygen = 32 g/mol =0.032 Kg/mol

now plugging these values to find T we get

`263.75=\sqrt{(3(8.314)(T))/(0.032)}`

solving the above equation we get

T= 89.24 K

Answer 2

T= 89.25 K

Step-by-step explanation:

Given that

V= 590 mph

We know that

1 mph = 0.44 m/s

That is why ,V= 263.75 m/s

We know that speed of the gas molecule is given as

`V=\sqrt{(3RT)/(M)}`

R= 8.314 J/mol.k

M= 32 g/mol = 0.032 kg/mol

T=Temperature in Kelvin unit

`V^2=(3RT)/(M)`

`T=(V^2* M)/(3R)`

`T=(263.76^2* 0.032)/(3* 8.314)\ K`

T= 89.25 K

Therefore the average temperature ,T = 89.25 K