The atomic mass of a nitrogen atom (N) is 14.0 u, while that of an oxygen atom (O) is 16.0 u. Three diatomic gases have the same temperature: nitrogen (N2), oxygen (O2), and nitric oxide (NO). Rank these gases in ascending order (smallest first), according to the values of their translational rms speeds.
A) N2, O2, NO
B) N2, NO, O2
C) O2, N2, NO
D) NO, N2, O2
Please explain step by step.
O2, NO, N2
To determine the translational rms speeds of the gases, we can use the root mean square speed formula:
v_rms = √(3RT/M)
where:
- v_rms is the translational rms 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
Step 1: Compare the molar masses of N2, O2, and NO.
- The molar mass of N2 is 2 × 14.0 u = 28.0 u.
- The molar mass of O2 is 2 × 16.0 u = 32.0 u.
- The molar mass of NO is 14.0 u + 16.0 u = 30.0 u.
Step 2: Since the temperature is the same for all gases, it can be canceled out in the comparison.
Step 3: Calculate the values of 3RT/M for each gas.
- For N2: (3 × 8.314 × T) / 28.0
- For O2: (3 × 8.314 × T) / 32.0
- For NO: (3 × 8.314 × T) / 30.0
As the temperature is the same for all the gases, we can ignore it in the comparison.
Step 4: Compare the values of 3RT/M for each gas.
- For N2: 8.314 / 28.0
- For O2: 8.314 / 32.0
- For NO: 8.314 / 30.0
Step 5: Simplify the values.
- For N2: 0.2970
- For O2: 0.2598
- For NO: 0.2771
Step 6: Rank the gases according to the values obtained in step 5, in ascending order (smallest first).
- N2: 0.2970
- NO: 0.2771
- O2: 0.2598
Therefore, the correct answer is B) N2, NO, O2.