Answer:
A) d. (1/4)(2000m/s) = 500 m/s
B) c. 4000 J
C) f. None of the above (2149.24 m/s)
Explanation:
A)
The translational kinetic energy of a gas molecule is given as:
K.E = (3/2)KT
where,
K = Boltzman's Constant = 1.38 x 1^-23 J/K
T = Absolute Temperature
but,
K.E = (1/2) mv²
where,
v = root mean square velocity
m = mass of one mole of a gas
Comparing both equations:
(3/2)KT = (1/2) mv²
v = √(3KT)/m _____ eqn (1)
<u>FOR HYDROGEN:</u>
v = √(3KT)/m = 2000 m/s _____ eqn (2)
<u>FOR OXYGEN:</u>
velocity of oxygen = √(3KT)/(mass of oxygen)
Here,
mass of 1 mole of oxygen = 16 m
velocity of oxygen = √(3KT)/(16 m)
velocity of oxygen = (1/4) √(3KT)/m
using eqn (2)
<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>
B)
K.E = (3/2)KT
Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.
K.E of Oxygen = K.E of Hydrogen
<u>K.E of Oxygen = 4000 J</u>
C)
using eqn (2)
At, T = 50°C = 323 k
v = √(3KT)/m = 2000 m/s
m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²
m = 3.343 x 10^-27 kg
So, now for this value of m and T = 100°C = 373 k
v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)
<u>v = 2149.24 m/s</u>
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