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At what temperature will the collision frequency in N2 be equal to that in He at 25°C if both gases are at a pressure of 1.00 bar collision diameter = 0.373 nm for N2 = 0.218 nm for He (Assume equal number of molecules per unit volume.)
a)46K
b)17.80 K
c)29.8 K
d)400K
Correct answer is option 'B'. Can you explain this answer?
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The collision frequency, denoted by Z, is given by the equation:

Z = (4 * N * π * d^2 * v) / (3 * V)

where:
- N is the number of molecules per unit volume
- d is the diameter of the molecule
- v is the root mean square velocity of the molecules
- V is the volume

To find the temperature at which the collision frequency in N2 is equal to that in He at 25°C, we need to equate the collision frequencies for the two gases.

Z(N2) = Z(He)

[(4 * N(N2) * π * d(N2)^2 * v(N2)] / (3 * V(N2)) = [(4 * N(He) * π * d(He)^2 * v(He)] / (3 * V(He))

Since the collision frequency depends on various factors, including the number of molecules, diameter, and velocity, we need specific values for these parameters for both N2 and He in order to calculate the temperature.
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At what temperature will the collision frequency in N2 be equal to that in He at 25°C if both gases are at a pressure of 1.00 bar collision diameter = 0.373 nm for N2 = 0.218 nm for He (Assume equal number of molecules per unit volume.)a)46Kb)17.80 Kc)29.8 Kd)400KCorrect answer is option 'B'. Can you explain this answer?
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At what temperature will the collision frequency in N2 be equal to that in He at 25°C if both gases are at a pressure of 1.00 bar collision diameter = 0.373 nm for N2 = 0.218 nm for He (Assume equal number of molecules per unit volume.)a)46Kb)17.80 Kc)29.8 Kd)400KCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about At what temperature will the collision frequency in N2 be equal to that in He at 25°C if both gases are at a pressure of 1.00 bar collision diameter = 0.373 nm for N2 = 0.218 nm for He (Assume equal number of molecules per unit volume.)a)46Kb)17.80 Kc)29.8 Kd)400KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At what temperature will the collision frequency in N2 be equal to that in He at 25°C if both gases are at a pressure of 1.00 bar collision diameter = 0.373 nm for N2 = 0.218 nm for He (Assume equal number of molecules per unit volume.)a)46Kb)17.80 Kc)29.8 Kd)400KCorrect answer is option 'B'. Can you explain this answer?.
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