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 in Kelvin) given the average velocity of CO₂, we can use the formula for average molecular speed:

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

Where:
- \( v \) = average velocity (in m/s)
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T \) = temperature in Kelvin
- \( m \) = molecular mass (in kg)

### Step 1: Convert the given average velocity from cm/s to m/s
The average velocity of CO₂ is given as \( 9 \times 10^4 \) cm/s. To convert this to m/s, we divide by 100:

\[ v = \frac{9 \times 10^4 \text{ cm/s}}{100} = 9 \times 10^2 \text{ m/s} \]

### Step 2: Determine the molecular mass of CO₂
The molecular mass of CO₂ can be calculated as follows:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol (and there are two oxygen atoms in CO₂)

Thus, the molecular mass of CO₂ is:

\[ m = 12 + 2 \times 16 = 12 + 32 = 44 \text{ g/mol} \]

To convert this to kg, we divide by 1000:

\[ m = \frac{44 \text{ g/mol}}{1000} = 0.044 \text{ kg/mol} \]

### Step 3: Substitute known values into the average velocity formula
Now we can substitute the values into the formula:

\[ 9 \times 10^2 = \sqrt{\frac{8 \times 8.314 \times T}{\pi \times 0.044}} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:

\[ (9 \times 10^2)^2 = \frac{8 \times 8.314 \times T}{\pi \times 0.044} \]

Calculating the left side:

\[ (9 \times 10^2)^2 = 81 \times 10^4 = 8100000 \]

### Step 5: Rearrange the equation to solve for T
Rearranging the equation gives:

\[ T = \frac{8100000 \times \pi \times 0.044}{8 \times 8.314} \]

### Step 6: Calculate the value of T
Using \( \pi \approx 3.14 \):

\[ T = \frac{8100000 \times 3.14 \times 0.044}{8 \times 8.314} \]

Calculating the numerator:

\[ 8100000 \times 3.14 \times 0.044 \approx 1.116 \times 10^6 \]

Calculating the denominator:

\[ 8 \times 8.314 \approx 66.512 \]

Now, dividing:

\[ T \approx \frac{1.116 \times 10^6}{66.512} \approx 16825.4 \text{ K} \]

### Final Answer
Thus, the value of T is approximately:

\[ T \approx 168.25 \text{ K} \]