Answer:
the temperature of the gas must have increased
Explanation:
As we know that
[tex]v = \sqrt{\frac{3RT}{M}}[/tex]
from this at room temperature the speed of air molecules is given as
[tex]v = \sqrt{\frac{3(8.31)(300)}{0.029}}[/tex]
[tex]v = 508 m/s[/tex]
now if the speed of the container was initially at
v = 2000 m/s
now we know that this speed is more than the speed of air molecules at room temperature
so here the temperature of the gas must have increased