If the universal gas constant is `8.3J//"mole"//K` and the Avogadro's number is `6xx10^(23)`, then the mean K.E. of the oxygen molecules at `327^(@)C` will be

Options

A. `2.49xx10^(20)J`
B. `4.10xx10^(-21)J`
C. `8.5xx10^(20)J`
D. `1.245xx10^(-20)J`

Explanation

To find the mean kinetic energy (K.E.) of oxygen molecules at a given temperature, we can use the formula:

\[ \text{Mean K.E.} = \frac{3}{2} \frac{R T}{N} \]

where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( N \) is Avogadro's number.

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature in Celsius is given as \( 327^\circ C \). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

Substituting the value:

\[ T = 327 + 273 = 600 \, K \]

### Step 2: Substitute the known values into the formula
We have:
- \( R = 8.3 \, \text{J/(mole K)} \)
- \( T = 600 \, K \)
- \( N = 6 \times 10^{23} \, \text{molecules/mole} \)

We can now substitute these values into the mean kinetic energy formula:

\[ \text{Mean K.E.} = \frac{3}{2} \frac{8.3 \times 600}{6 \times 10^{23}} \]

### Step 3: Calculate the numerator
First, calculate the numerator:

\[ 8.3 \times 600 = 4980 \, \text{J/mole} \]

### Step 4: Substitute the numerator back into the equation
Now, substitute this value back into the mean kinetic energy formula:

\[ \text{Mean K.E.} = \frac{3}{2} \frac{4980}{6 \times 10^{23}} \]

### Step 5: Calculate the fraction
Now calculate the fraction:

\[ \frac{4980}{6 \times 10^{23}} = \frac{4980}{6} \times 10^{-23} = 830 \times 10^{-23} = 8.3 \times 10^{-21} \, \text{J} \]

### Step 6: Multiply by \(\frac{3}{2}\)
Now multiply by \(\frac{3}{2}\):

\[ \text{Mean K.E.} = \frac{3}{2} \times 8.3 \times 10^{-21} = 12.45 \times 10^{-21} \, \text{J} \]

### Step 7: Final result
Thus, the mean kinetic energy of the oxygen molecules at \( 327^\circ C \) is:

\[ \text{Mean K.E.} = 1.245 \times 10^{-20} \, \text{J} \]