I sometimes do a lecture demonstration where I open a bottle of concentrated amm
ID: 963218 • Letter: I
Question
I sometimes do a lecture demonstration where I open a bottle of concentrated ammonia and time how long it takes for students in the row to smell the ammonia gas. I then measure the distance from students. Typical data are distance = 3.0 m; time = 15 seconds, Use the data to calculate the speed of the ammonia molecules in m/s. Calculate the speed of the ammonia molecules (ammonia is NH_3), in m/s, using the equation for squareroot mean square speed. Assume room temperature is 300 K. Why don't the answers from (a) and (b) agreeExplanation / Answer
a)Speed of ammonia using observation data:
Given: Distance = 3.0 m and time =15 s
Speed = Distance / Time
Speed of ammonia = 3.0 / 15 = 0.2 m/s
Hence speed of Ammonia is 0.2 m/s.
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b)Speed of ammonia using equation for mean square speed(u)
Formula for root mean square speed (u)
u = [3RT/M]1/2
Where,
R = molar gas constant = 8.314 J.K-1.mol-1.
T = 300 K
M = molar mass of ammonia (NH3) = 17 g = 17 x 10-3 Kg.
u = [(3 x 8.314 x 300) / 17 x 10-3]1/2
u = 663.4 m/s.
root mean square speed of ammonia is 663.4 m/s.
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c)Speed calculated based on observation is much small than the root mean square speed of ammonia, because
i)While calculating root mean square speed (RMS speed) we assume ideal behavior of ammonia i.e. no molecular association, no attractive interaction among molecules etc.
ii) Ammonia molecules before reaching to nose undergo very large number of collisins. These collision scatter them in all possible direction and takes more time to reach the nose.
iii)While calculating speed on the basis of observation i. (a) we assume that molecules trace just a linear path but its not the case at all. Molecules do scatter in all possible direction and traces acutally long path to reach nose.
And hence the difference between two speeds.
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