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 solve the problem, we will use the formula for the average speed of gas molecules, which is given by:

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

Where:
- \( v \) is the average speed of the gas,
- \( R \) 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.

Given:
\[ v = \sqrt{\frac{28}{88}} \times 10^3 \, \text{m/s} \]

### Step 1: Square both sides of the equation
We start by squaring both sides of the equation to eliminate the square root:

\[ v^2 = \frac{8RT}{\pi M} \]

Substituting the expression for \( v \):

\[ \left(\sqrt{\frac{28}{88}} \times 10^3\right)^2 = \frac{8RT}{\pi M} \]

### Step 2: Calculate \( v^2 \)
Now we calculate \( v^2 \):

\[ v^2 = \frac{28}{88} \times 10^6 \]

### Step 3: Rearranging the equation
Rearranging the equation to solve for \( T \):

\[ T = \frac{v^2 \cdot \pi M}{8R} \]

### Step 4: Substitute values
Now we substitute the known values into the equation. The molar mass \( M \) of \( CH_4 \) is 16 g/mol, which we convert to kg/mol:

\[ M = 16 \times 10^{-3} \, \text{kg/mol} = 0.016 \, \text{kg/mol} \]

Now substituting \( R = 8.314 \, \text{J/(mol·K)} \):

\[ T = \frac{\left(\frac{28}{88} \times 10^6\right) \cdot \pi \cdot 0.016}{8 \cdot 8.314} \]

### Step 5: Calculate \( T \)
Calculating the right-hand side:

1. Calculate \( \frac{28}{88} \):
\[ \frac{28}{88} = 0.3182 \]

2. Calculate \( 0.3182 \times 10^6 \):
\[ 0.3182 \times 10^6 = 318200 \]

3. Now substitute into the equation:
\[ T = \frac{318200 \cdot \pi \cdot 0.016}{8 \cdot 8.314} \]

4. Calculate \( \pi \cdot 0.016 \):
\[ \pi \cdot 0.016 \approx 0.05027 \]

5. Calculate \( 8 \cdot 8.314 \):
\[ 8 \cdot 8.314 = 66.512 \]

6. Now we can calculate \( T \):
\[ T = \frac{318200 \cdot 0.05027}{66.512} \approx 240.55 \, \text{K} \]

### Step 6: Convert Kelvin to Celsius
To convert Kelvin to Celsius, we use the formula:

\[ T(°C) = T(K) - 273 \]

So:

\[ T(°C) = 240.55 - 273 \approx -32.45 °C \]

### Final Answer:
The value of \( T \) is approximately \(-32.45 °C\).

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