The window has a surface of 1.5 m2*... *The compartment has a volume of 12 m3*... *The train speeds 80 kph to the west*.The south wind is blowing at 5 mps*...*How quickly will the space be refreshed?*

Train speeds at 80kph i.e 22m/s
The train is moving towards west (22m/s) and wind to south which is (5m/s)

 So resultant wind through window
 √((22)^2 + (5)^2)
Which is a 13m/sA
At angle of {tan^-1(5/22+5)} which is 10.49 degrees.

Assuming Window is fully open
So 13m/s multiplied by dimensions of window
Then  13m/s × 1.5 cu. m = 19.5 cu.m

But the dimensions of compartment is a 12 cu.m ....so 19.5/12 = 1.625....so the space will be refreshed in 1.625 seconds

Aniket Kulkarni