What is kinetic energy of 1 gm of `O_(2)` at `47^(@)C` ?

Options

A. `2.17 xx 10^(2) J`
B. `2.24 xx 10^(2) J`
C. `1.24 xx 10^(2) J`
D. None of these

Explanation

To find the kinetic energy of 1 gram of \( O_2 \) at \( 47^\circ C \), we can follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin.
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Substituting the given temperature:
\[ T(K) = 47 + 273 = 320 \, K \]

### Step 2: Calculate the number of moles of \( O_2 \).
The number of moles \( n \) can be calculated using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
Given that the mass of \( O_2 \) is 1 gram and the molar mass of \( O_2 \) is 32 g/mol:
\[ n = \frac{1 \, \text{g}}{32 \, \text{g/mol}} = \frac{1}{32} \, \text{mol} \]

### Step 3: Use the kinetic energy formula.
The kinetic energy \( KE \) of an ideal gas can be calculated using the formula:
\[ KE = \frac{3}{2} nRT \]
Where:
- \( R \) is the gas constant, \( R = 8.314 \, \text{J/(mol K)} \)
- \( T \) is the temperature in Kelvin

Substituting the values we have:
\[ KE = \frac{3}{2} \left(\frac{1}{32}\right) (8.314) (320) \]

### Step 4: Calculate the kinetic energy.
Now, we can calculate the kinetic energy:
\[ KE = \frac{3}{2} \times \frac{1}{32} \times 8.314 \times 320 \]
Calculating the multiplication:
\[ KE = \frac{3 \times 8.314 \times 320}{2 \times 32} \]
\[ KE = \frac{3 \times 8.314 \times 320}{64} \]
\[ KE = \frac{7987.68}{64} \approx 124.55 \, \text{J} \]

Thus, the kinetic energy of 1 gram of \( O_2 \) at \( 47^\circ C \) is approximately \( 124.55 \, \text{J} \).

### Final Answer:
The kinetic energy of 1 gram of \( O_2 \) at \( 47^\circ C \) is approximately \( 124.55 \, \text{J} \).

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