Answer:
T= 89.25 K
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
[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]
R= 8.314 J/mol.k
M= 32 g/mol = 0.032 kg/mol
T=Temperature in Kelvin unit
[tex]V^2=\dfrac{3RT}{M}[/tex]
[tex]T=\dfrac{V^2\times M}{3R}[/tex]
[tex]T=\dfrac{263.76^2\times 0.032}{3\times 8.314}\ K[/tex]
T= 89.25 K
Therefore the average temperature ,T = 89.25 K
Answer:
temperature does the average speed of an oxygen molecule equal that of an airplane moving at 590 mph = 89.24 K
Explanation:
Average speed of oxygen molecule is given by
[tex]v= \sqrt{\frac{3RT}{M} }[/tex]
[tex]590\times0.44704 = 263.75[/tex] 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
[tex]263.75=\sqrt{\frac{3(8.314)(T)}{0.032}}[/tex]
solving the above equation we get
T= 89.24 K
