The average speed at temperature `T^(@)C` of `CH_(4)(g)` is `sqrt((28)/(88))xx10^(3)ms^(-1)`. What is the value of T ?

Options

A. `240.55^(@)C`
B. `-32.45^(@)C`
C. `3000^(@)C`
D. `-24.055^(@)C`

Explanation

To find the temperature \( T \) given the average speed of methane \( CH_4(g) \) at temperature \( T \) in degrees Celsius, we can use the formula for the average speed of a gas:

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

Where:
- \( v \) is the average speed,
- \( R \) is the universal gas constant (8.314 J/(mol·K)),
- \( T \) is the temperature in Kelvin,
- \( m \) is the molar mass of the gas in kg/mol.

### Step-by-Step Solution:

1. **Set up the equation**:
We know from the problem that the average speed \( v \) is given by:
\[ v = \sqrt{\frac{28}{88}} \times 10^3 \, \text{m/s} \]
Therefore, we can equate this to the formula for average speed:
\[ \sqrt{\frac{8RT}{\pi m}} = \sqrt{\frac{28}{88}} \times 10^3 \]

2. **Square both sides**:
Squaring both sides to eliminate the square root gives:
\[ \frac{8RT}{\pi m} = \frac{28}{88} \times 10^6 \]

3. **Substitute known values**:
The molar mass of methane \( CH_4 \) is 16 g/mol, which is 0.016 kg/mol. Thus, we can substitute \( m = 0.016 \, \text{kg/mol} \) into the equation:
\[ \frac{8 \times 8.314 \times T}{\pi \times 0.016} = \frac{28}{88} \times 10^6 \]

4. **Calculate the left side**:
First, calculate \( \pi \) (approximately 3.14):
\[ \frac{8 \times 8.314 \times T}{3.14 \times 0.016} \]
This simplifies to:
\[ \frac{66.512 \times T}{0.05024} \]
Which is approximately:
\[ 1324.95 \times T \]

5. **Set the equation**:
Now we have:
\[ 1324.95 \times T = \frac{28 \times 10^6}{88} \]

6. **Calculate the right side**:
Calculate \( \frac{28 \times 10^6}{88} \):
\[ \frac{28 \times 10^6}{88} = 318181.82 \]

7. **Solve for \( T \)**:
Now we can solve for \( T \):
\[ T = \frac{318181.82}{1324.95} \approx 240.0 \, \text{K} \]

8. **Convert to Celsius**:
To convert Kelvin to Celsius:
\[ T_{C} = T_{K} - 273.15 \approx 240.0 - 273.15 \approx -33.15 \, ^\circ C \]

### Final Answer:
The value of \( T \) is approximately -33.15 °C.