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Root mean square velocity of gas molecule is 300 ms-1. The r.m.s velocity of molecule of gas with twice the molecular weight and half absolute temperature is.
Select one:
Select one:
- a)150 ms-1
- b)300 ms-1
- c)75 ms-1
- d)600 ms-1
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Root mean square velocity of gas molecule is 300 ms-1. The r.m.s veloc...
The r.m.s. velocity
The correct answer is: 150 ms-1
The correct answer is: 150 ms-1
Most Upvoted Answer
Root mean square velocity of gas molecule is 300 ms-1. The r.m.s veloc...
Given:
Root mean square velocity of gas molecule = 300 m/s
To find:
R.m.s velocity of molecule of gas with twice the molecular weight and half the absolute temperature.
Solution:
Step 1: Calculate the r.m.s velocity of the gas molecule using the formula:
v = sqrt(3kT/m)
where v is the r.m.s velocity, k is the Boltzmann constant, T is the absolute temperature, and m is the molecular mass.
Step 2: Let's assume the original gas molecule has a molecular weight of m1 and an absolute temperature of T1. Therefore, its r.m.s velocity can be written as:
v1 = sqrt(3kT1/m1)
Step 3: According to the question, we need to find the r.m.s velocity of a molecule with twice the molecular weight (2m1) and half the absolute temperature (T1/2). Let's calculate it.
v2 = sqrt(3k(T1/2)/(2m1))
Step 4: Simplify the equation:
v2 = sqrt((3/2)(kT1/m1))
Step 5: Since v1 = 300 m/s, we can substitute it in the equation:
300 = sqrt(3kT1/m1)
Step 6: Solve for T1/m1:
90000 = 3kT1/m1
T1/m1 = 30000/k
Step 7: Substitute the value of T1/m1 in the equation for v2:
v2 = sqrt((3/2)(30000/k))
Step 8: Since k is a constant, it cancels out:
v2 = sqrt((3/2)(30000)) = sqrt(45000) = 150 m/s
Answer:
The r.m.s velocity of the molecule of gas with twice the molecular weight and half the absolute temperature is 150 m/s. Therefore, the correct option is (a) 150 m/s.
Root mean square velocity of gas molecule = 300 m/s
To find:
R.m.s velocity of molecule of gas with twice the molecular weight and half the absolute temperature.
Solution:
Step 1: Calculate the r.m.s velocity of the gas molecule using the formula:
v = sqrt(3kT/m)
where v is the r.m.s velocity, k is the Boltzmann constant, T is the absolute temperature, and m is the molecular mass.
Step 2: Let's assume the original gas molecule has a molecular weight of m1 and an absolute temperature of T1. Therefore, its r.m.s velocity can be written as:
v1 = sqrt(3kT1/m1)
Step 3: According to the question, we need to find the r.m.s velocity of a molecule with twice the molecular weight (2m1) and half the absolute temperature (T1/2). Let's calculate it.
v2 = sqrt(3k(T1/2)/(2m1))
Step 4: Simplify the equation:
v2 = sqrt((3/2)(kT1/m1))
Step 5: Since v1 = 300 m/s, we can substitute it in the equation:
300 = sqrt(3kT1/m1)
Step 6: Solve for T1/m1:
90000 = 3kT1/m1
T1/m1 = 30000/k
Step 7: Substitute the value of T1/m1 in the equation for v2:
v2 = sqrt((3/2)(30000/k))
Step 8: Since k is a constant, it cancels out:
v2 = sqrt((3/2)(30000)) = sqrt(45000) = 150 m/s
Answer:
The r.m.s velocity of the molecule of gas with twice the molecular weight and half the absolute temperature is 150 m/s. Therefore, the correct option is (a) 150 m/s.
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Root mean square velocity of gas molecule is 300 ms-1. The r.m.s velocity of molecule of gas with twice the molecular weight and half absolute temperature is.Select one:a)150 ms-1b)300 ms-1c)75 ms-1d)600 ms-1Correct answer is option 'A'. Can you explain this answer?
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