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At what temperature the root mean square velocity of O2 gas molecules will be 75% of root mean square velocity of H2 gas molecule and 200 k?
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At what temperature the root mean square velocity of O2 gas molecules ...
Understanding Root Mean Square Velocity
The root mean square (RMS) velocity of a gas is given by the formula:
RMS velocity = √(3RT/M)
Where:
- R = Universal gas constant
- T = Temperature in Kelvin
- M = Molar mass of the gas
Given Information
- RMS velocity of O2 should be 75% of the RMS velocity of H2 at a temperature of 200 K.
Molar Masses
- Molar mass of H2 = 2 g/mol
- Molar mass of O2 = 32 g/mol
RMS Velocity Relation
Let v_H2 be the RMS velocity of H2 and v_O2 be the RMS velocity of O2.
- v_H2 = √(3RT_H2/M_H2)
- v_O2 = √(3RT_O2/M_O2)
According to the problem:
v_O2 = 0.75 * v_H2
Substituting the RMS expressions:
√(3RT_O2/M_O2) = 0.75 * √(3RT_H2/M_H2)
Temperature Calculation
Since we need the temperature for O2 (T_O2) to satisfy the above equation at T_H2 = 200 K:
1. Substituting the values of molar masses:
√(3RT_O2/32) = 0.75 * √(3R * 200/2)
2. Simplifying the equation:
√(T_O2/32) = 0.75 * √(200/2)
3. Solving for T_O2:
T_O2 = 32 * (0.75 * √(100))^2
This calculation gives:
T_O2 = 32 * (0.75^2 * 100) = 32 * 56.25 = 1800 K
Conclusion
Thus, the temperature at which the RMS velocity of O2 is 75% of that of H2 at 200 K is approximately 1800 K.
The root mean square (RMS) velocity of a gas is given by the formula:
RMS velocity = √(3RT/M)
Where:
- R = Universal gas constant
- T = Temperature in Kelvin
- M = Molar mass of the gas
Given Information
- RMS velocity of O2 should be 75% of the RMS velocity of H2 at a temperature of 200 K.
Molar Masses
- Molar mass of H2 = 2 g/mol
- Molar mass of O2 = 32 g/mol
RMS Velocity Relation
Let v_H2 be the RMS velocity of H2 and v_O2 be the RMS velocity of O2.
- v_H2 = √(3RT_H2/M_H2)
- v_O2 = √(3RT_O2/M_O2)
According to the problem:
v_O2 = 0.75 * v_H2
Substituting the RMS expressions:
√(3RT_O2/M_O2) = 0.75 * √(3RT_H2/M_H2)
Temperature Calculation
Since we need the temperature for O2 (T_O2) to satisfy the above equation at T_H2 = 200 K:
1. Substituting the values of molar masses:
√(3RT_O2/32) = 0.75 * √(3R * 200/2)
2. Simplifying the equation:
√(T_O2/32) = 0.75 * √(200/2)
3. Solving for T_O2:
T_O2 = 32 * (0.75 * √(100))^2
This calculation gives:
T_O2 = 32 * (0.75^2 * 100) = 32 * 56.25 = 1800 K
Conclusion
Thus, the temperature at which the RMS velocity of O2 is 75% of that of H2 at 200 K is approximately 1800 K.
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At what temperature the root mean square velocity of O2 gas molecules will be 75% of root mean square velocity of H2 gas molecule and 200 k?
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At what temperature the root mean square velocity of O2 gas molecules will be 75% of root mean square velocity of H2 gas molecule and 200 k? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At what temperature the root mean square velocity of O2 gas molecules will be 75% of root mean square velocity of H2 gas molecule and 200 k? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At what temperature the root mean square velocity of O2 gas molecules will be 75% of root mean square velocity of H2 gas molecule and 200 k?.
At what temperature the root mean square velocity of O2 gas molecules will be 75% of root mean square velocity of H2 gas molecule and 200 k? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At what temperature the root mean square velocity of O2 gas molecules will be 75% of root mean square velocity of H2 gas molecule and 200 k? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At what temperature the root mean square velocity of O2 gas molecules will be 75% of root mean square velocity of H2 gas molecule and 200 k?.
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