The speed of the atom in Question 03.2 is equal to $c_{rms}$, the root mean square speed of the atoms of the gas in the oven - AQA - A-Level Physics - Question 3 - 2021 - Paper 2
Question 3
The speed of the atom in Question 03.2 is equal to $c_{rms}$, the root mean square speed of the atoms of the gas in the oven.
The molar mass of the gas is 0.209 kg ... show full transcript
Worked Solution & Example Answer:The speed of the atom in Question 03.2 is equal to $c_{rms}$, the root mean square speed of the atoms of the gas in the oven - AQA - A-Level Physics - Question 3 - 2021 - Paper 2
Step 1
The speed of the atom in Question 03.2 is equal to $c_{rms}$
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Answer
To calculate the root mean square speed, we use the equation:
c_{rms} = rac{3RT}{M}
Where:
R is the universal gas constant, approximately 8.31extJmol−1extK−1.
T is the temperature in Kelvin.
M is the molar mass in kg mol−1.
Rearranging gives:
T = rac{c_{rms} imes M}{3R}
Step 2
Substituting values for $c_{rms}$
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Answer
From the previous calculations, we take:
crms=500extms−1
M=0.209extkgmol−1
Substituting these values into the equation:
T = rac{500 ext{ m s}^{-1} imes 0.209 ext{ kg mol}^{-1}}{3 imes 8.31 ext{ J mol}^{-1} ext{ K}^{-1}}
Step 3
Calculating the temperature
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Answer
Calculating the above expression:
First, calculate the numerator: 500imes0.209=104.5.
Then calculate the denominator: 3imes8.31=24.93.
Thus:
T = rac{104.5}{24.93} \\ T \\approx 4.19 ext{ K}
Therefore, the temperature is approximately T=1930extK. This shows that the atoms in the oven are at a temperature of roughly 1930 Kelvin.
