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At 400 K, the root mean square (rms) speed of a gas X (molecule weight = 40) is equal is equal to the most probable speed of gas Y at 60 K. The mo lecular weight of the gas Y is:
Correct answer is '1.25'. Can you explain this answer?
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At 400 K, the root mean square (rms) speed of a gas X (molecule weight...
**Answer:**
To solve this problem, we need to use the relationship between the root mean square (rms) speed and the most probable speed of gas molecules.
The root mean square (rms) speed of gas molecules can be calculated using the formula:
**v_rms = √(3RT/M)**
Where:
- v_rms is the root mean square speed
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- M is the molar mass of the gas in kg/mol
The most probable speed of gas molecules can be calculated using the formula:
**v_mostprobable = √(2RT/M)**
Where:
- v_mostprobable is the most probable speed
We are given that the rms speed of gas X at 400 K is equal to the most probable speed of gas Y at 60 K. Let's denote the molar mass of gas Y as M_Y.
Using the given information, we can set up the following equation:
**v_rms(X) = v_mostprobable(Y)**
**√(3RT_X/M_X) = √(2RT_Y/M_Y)**
Squaring both sides of the equation, we get:
**3RT_X/M_X = 2RT_Y/M_Y**
Now we can substitute in the given values and solve for M_Y:
**3R(400 K)/40 = 2R(60 K)/M_Y**
Simplifying the equation:
**3(8.314 J/(mol·K))(400 K)/40 = 2(8.314 J/(mol·K))(60 K)/M_Y**
**2494.2 J/mol = 997.04 J/mol/M_Y**
Dividing both sides by 997.04 J/mol, we get:
**M_Y = 997.04 J/mol / 2494.2 J/mol**
**M_Y = 0.4 mol**
Converting mol to g/mol:
**M_Y = 0.4 mol × 1000 g/mol**
**M_Y = 400 g/mol**
Therefore, the molar mass of gas Y is 400 g/mol, which is equivalent to 4 times the molar mass of gas X (40 g/mol).
To solve this problem, we need to use the relationship between the root mean square (rms) speed and the most probable speed of gas molecules.
The root mean square (rms) speed of gas molecules can be calculated using the formula:
**v_rms = √(3RT/M)**
Where:
- v_rms is the root mean square speed
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- M is the molar mass of the gas in kg/mol
The most probable speed of gas molecules can be calculated using the formula:
**v_mostprobable = √(2RT/M)**
Where:
- v_mostprobable is the most probable speed
We are given that the rms speed of gas X at 400 K is equal to the most probable speed of gas Y at 60 K. Let's denote the molar mass of gas Y as M_Y.
Using the given information, we can set up the following equation:
**v_rms(X) = v_mostprobable(Y)**
**√(3RT_X/M_X) = √(2RT_Y/M_Y)**
Squaring both sides of the equation, we get:
**3RT_X/M_X = 2RT_Y/M_Y**
Now we can substitute in the given values and solve for M_Y:
**3R(400 K)/40 = 2R(60 K)/M_Y**
Simplifying the equation:
**3(8.314 J/(mol·K))(400 K)/40 = 2(8.314 J/(mol·K))(60 K)/M_Y**
**2494.2 J/mol = 997.04 J/mol/M_Y**
Dividing both sides by 997.04 J/mol, we get:
**M_Y = 997.04 J/mol / 2494.2 J/mol**
**M_Y = 0.4 mol**
Converting mol to g/mol:
**M_Y = 0.4 mol × 1000 g/mol**
**M_Y = 400 g/mol**
Therefore, the molar mass of gas Y is 400 g/mol, which is equivalent to 4 times the molar mass of gas X (40 g/mol).
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At 400 K, the root mean square (rms) speed of a gas X (molecule weight...
M= 4
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At 400 K, the root mean square (rms) speed of a gas X (molecule weight = 40) is equal is equal to the most probable speed of gas Y at 60 K. The mo lecular weight of the gas Y is:Correct answer is '1.25'. Can you explain this answer?
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At 400 K, the root mean square (rms) speed of a gas X (molecule weight = 40) is equal is equal to the most probable speed of gas Y at 60 K. The mo lecular weight of the gas Y is:Correct answer is '1.25'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about At 400 K, the root mean square (rms) speed of a gas X (molecule weight = 40) is equal is equal to the most probable speed of gas Y at 60 K. The mo lecular weight of the gas Y is:Correct answer is '1.25'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 400 K, the root mean square (rms) speed of a gas X (molecule weight = 40) is equal is equal to the most probable speed of gas Y at 60 K. The mo lecular weight of the gas Y is:Correct answer is '1.25'. Can you explain this answer?.
At 400 K, the root mean square (rms) speed of a gas X (molecule weight = 40) is equal is equal to the most probable speed of gas Y at 60 K. The mo lecular weight of the gas Y is:Correct answer is '1.25'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about At 400 K, the root mean square (rms) speed of a gas X (molecule weight = 40) is equal is equal to the most probable speed of gas Y at 60 K. The mo lecular weight of the gas Y is:Correct answer is '1.25'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 400 K, the root mean square (rms) speed of a gas X (molecule weight = 40) is equal is equal to the most probable speed of gas Y at 60 K. The mo lecular weight of the gas Y is:Correct answer is '1.25'. Can you explain this answer?.
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