The average velocity of `CO_(2)` at T K is `9 xx 10^(4) cm s^(-1)`. The value of T is

Options

A. 1682.5K
B. 2143.37
C. 373.20K
D. 1240 K

Answer

To find the value of T (temperature) given the average velocity of CO₂, we can use the formula for the average velocity of gas molecules:

\[ V = \sqrt{\frac{8RT}{\pi m}} \]

Where:
- \( V \) = average velocity
- \( R \) = gas constant
- \( T \) = temperature in Kelvin
- \( m \) = molar mass of the gas

### Step-by-step Solution:

1. **Identify the given values**:
- Average velocity \( V = 9 \times 10^4 \, \text{cm/s} \)
- Gas constant \( R = 8.314 \, \text{J/(mol K)} \) (Note: Convert to appropriate units if necessary)
- Molar mass of CO₂ \( m = 44 \, \text{g/mol} \)

2. **Rearrange the formula to solve for T**:
\[ V^2 = \frac{8RT}{\pi m} \]
Rearranging gives:
\[ T = \frac{V^2 \cdot \pi m}{8R} \]

3. **Substitute the known values into the equation**:
\[ T = \frac{(9 \times 10^4)^2 \cdot \pi \cdot 44}{8 \cdot 8.314} \]

4. **Calculate \( V^2 \)**:
\[ V^2 = (9 \times 10^4)^2 = 8.1 \times 10^9 \, \text{cm}^2/\text{s}^2 \]

5. **Substitute \( V^2 \) back into the equation**:
\[ T = \frac{8.1 \times 10^9 \cdot 3.14 \cdot 44}{8 \cdot 8.314} \]

6. **Calculate the numerator**:
\[ \text{Numerator} = 8.1 \times 10^9 \cdot 3.14 \cdot 44 \approx 1.12 \times 10^{12} \]

7. **Calculate the denominator**:
\[ \text{Denominator} = 8 \cdot 8.314 \approx 66.512 \]

8. **Divide the numerator by the denominator**:
\[ T \approx \frac{1.12 \times 10^{12}}{66.512} \approx 16,182.55 \, \text{K} \]

### Final Answer:
The value of \( T \) is approximately **16,182.55 K**.