To find the root mean square (RMS) velocity of ethane at the same temperature as carbon dioxide, we can follow these steps:
### Step 1: Understand the formula for RMS velocity
The RMS velocity (v_rms) is given by the formula:
\[
v_{\text{rms}} = \sqrt{\frac{3RT}{M}}
\]
where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( M \) is the molar mass of the gas in kg/mol.
### Step 2: Calculate the molar mass of carbon dioxide (CO₂)
The molar mass of CO₂ can be calculated as follows:
- Carbon (C) has a molar mass of approximately 12 g/mol, and there is 1 carbon atom.
- Oxygen (O) has a molar mass of approximately 16 g/mol, and there are 2 oxygen atoms.
Thus, the molar mass of CO₂ is:
\[
M_{\text{CO}_2} = 12 + (2 \times 16) = 12 + 32 = 44 \text{ g/mol} = 0.044 \text{ kg/mol}
\]
### Step 3: Calculate the molar mass of ethane (C₂H₆)
The molar mass of ethane can be calculated as follows:
- Carbon (C) has a molar mass of approximately 12 g/mol, and there are 2 carbon atoms.
- Hydrogen (H) has a molar mass of approximately 1 g/mol, and there are 6 hydrogen atoms.
Thus, the molar mass of ethane is:
\[
M_{\text{C}_2\text{H}_6} = (2 \times 12) + (6 \times 1) = 24 + 6 = 30 \text{ g/mol} = 0.030 \text{ kg/mol}
\]
### Step 4: Set up the ratio of RMS velocities
Since we are comparing the RMS velocities of CO₂ and ethane at the same temperature, we can use the ratio of their RMS velocities:
\[
\frac{v_{\text{rms, C}_2\text{H}_6}}{v_{\text{rms, CO}_2}} = \sqrt{\frac{M_{\text{CO}_2}}{M_{\text{C}_2\text{H}_6}}}
\]
### Step 5: Substitute known values
We know:
- \( v_{\text{rms, CO}_2} = 4.4 \times 10^4 \text{ cm/s} \)
- \( M_{\text{CO}_2} = 44 \text{ g/mol} \)
- \( M_{\text{C}_2\text{H}_6} = 30 \text{ g/mol} \)
Substituting these values into the ratio:
\[
\frac{v_{\text{rms, C}_2\text{H}_6}}{4.4 \times 10^4} = \sqrt{\frac{44}{30}}
\]
### Step 6: Calculate the square root
Calculating the square root:
\[
\sqrt{\frac{44}{30}} = \sqrt{1.4667} \approx 1.21
\]
### Step 7: Solve for \( v_{\text{rms, C}_2\text{H}_6} \)
Now, we can find \( v_{\text{rms, C}_2\text{H}_6} \):
\[
v_{\text{rms, C}_2\text{H}_6} = 4.4 \times 10^4 \times 1.21 \approx 5.33 \times 10^4 \text{ cm/s}
\]
### Final Answer
The RMS velocity of ethane at the same temperature is approximately:
\[
v_{\text{rms, C}_2\text{H}_6} \approx 5.33 \times 10^4 \text{ cm/s}
\]
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