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The average velocity of anideal gas molecule at 27°C is 0.3 m/s. The average velocity at 927°C will be:
- a)0.6 m/s
- b)0.3 m/s
- c)0.9 m/s
- d)3.0 m/s
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The average velocity of anideal gas molecule at 27°C is 0.3 m/s. ...
We know,
V1/V2 = [T1/T2]1/2
T1 = 27oC
= 300K
T2 = 927oC
= 1200K
V1 = 0.3ms-1
V2 = [T2/T1]1/2 × V1
V1/V2 = [T1/T2]1/2
T1 = 27oC
= 300K
T2 = 927oC
= 1200K
V1 = 0.3ms-1
V2 = [T2/T1]1/2 × V1
= [1200/300] × 0.3
= 0.6ms-1
= 0.6ms-1
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The average velocity of anideal gas molecule at 27°C is 0.3 m/s. ...
The average velocity of an ideal gas molecule at 27°C can be calculated using the root mean square velocity formula:
v = √(3kT/m)
where:
v = average velocity
k = Boltzmann constant (1.38 × 10^-23 J/K)
T = temperature in Kelvin (27°C = 300 K)
m = mass of the gas molecule
The mass of the gas molecule will depend on the specific gas being considered. For example, the average velocity of a nitrogen molecule (N2) at 27°C can be calculated using its molar mass:
m(N2) = 28 g/mol
Converting the mass to kilograms:
m(N2) = 28 × 10^-3 kg/mol
The average velocity of a nitrogen molecule at 27°C would then be:
v(N2) = √(3 × 1.38 × 10^-23 J/K × 300 K / (28 × 10^-3 kg/mol))
Calculating this value would give you the average velocity of a nitrogen molecule at 27°C. The specific average velocity will vary for different gases depending on their molar mass.
v = √(3kT/m)
where:
v = average velocity
k = Boltzmann constant (1.38 × 10^-23 J/K)
T = temperature in Kelvin (27°C = 300 K)
m = mass of the gas molecule
The mass of the gas molecule will depend on the specific gas being considered. For example, the average velocity of a nitrogen molecule (N2) at 27°C can be calculated using its molar mass:
m(N2) = 28 g/mol
Converting the mass to kilograms:
m(N2) = 28 × 10^-3 kg/mol
The average velocity of a nitrogen molecule at 27°C would then be:
v(N2) = √(3 × 1.38 × 10^-23 J/K × 300 K / (28 × 10^-3 kg/mol))
Calculating this value would give you the average velocity of a nitrogen molecule at 27°C. The specific average velocity will vary for different gases depending on their molar mass.
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The average velocity of anideal gas molecule at 27°C is 0.3 m/s. The average velocity at 927°C will be:a)0.6 m/sb)0.3 m/sc)0.9 m/sd)3.0 m/sCorrect answer is option 'D'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The average velocity of anideal gas molecule at 27°C is 0.3 m/s. The average velocity at 927°C will be:a)0.6 m/sb)0.3 m/sc)0.9 m/sd)3.0 m/sCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The average velocity of anideal gas molecule at 27°C is 0.3 m/s. The average velocity at 927°C will be:a)0.6 m/sb)0.3 m/sc)0.9 m/sd)3.0 m/sCorrect answer is option 'D'. Can you explain this answer?.
The average velocity of anideal gas molecule at 27°C is 0.3 m/s. The average velocity at 927°C will be:a)0.6 m/sb)0.3 m/sc)0.9 m/sd)3.0 m/sCorrect answer is option 'D'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The average velocity of anideal gas molecule at 27°C is 0.3 m/s. The average velocity at 927°C will be:a)0.6 m/sb)0.3 m/sc)0.9 m/sd)3.0 m/sCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The average velocity of anideal gas molecule at 27°C is 0.3 m/s. The average velocity at 927°C will be:a)0.6 m/sb)0.3 m/sc)0.9 m/sd)3.0 m/sCorrect answer is option 'D'. Can you explain this answer?.
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