The pressure exerted by an ideal gas is `P = (1)/(3) (M)/(V)C^(2)`, where the symbols have their usual meaning. Using standard gas equation, PV = RT, we find that `C^(2) = (3 RT)/(M) or C^(2) oo T`. Average kinetic energy of translation of one mole of gas ` =(1)/(2) MC^(2) = (3 RT)/(2)` with the help of the passage given above, choose the most appropriate alternative for each of the following quetions :
Average thermal energy of a helium atom at 600 K would be

Options

A. `6.21 xx 10^(-21) J`
B. `1.24 xx 10^(-20) J`
C. `1.24 xx 10^(-21) J`
D. `1.24 xx 10^(21) J`

Explanation

To find the average thermal energy of a helium atom at 600 K, we can follow these steps:

### Step 1: Understand the relationship between average kinetic energy and temperature
The average kinetic energy (KE) of one mole of an ideal gas is given by the formula:
\[ KE = \frac{3}{2} RT \]
where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.

### Step 2: Convert the average kinetic energy per mole to average kinetic energy per atom
To find the average kinetic energy per atom, we can use the relationship:
\[ KE_{\text{atom}} = \frac{KE_{\text{mole}}}{N} \]
where \( N \) is Avogadro's number (\( N \approx 6.022 \times 10^{23} \) mol\(^{-1}\)). Thus,
\[ KE_{\text{atom}} = \frac{3}{2} \frac{RT}{N} \]

### Step 3: Substitute the value of Boltzmann's constant
We know that the Boltzmann constant \( k \) is related to the universal gas constant \( R \) by:
\[ k = \frac{R}{N} \]
Therefore, we can rewrite the average kinetic energy per atom as:
\[ KE_{\text{atom}} = \frac{3}{2} k T \]

### Step 4: Calculate the average kinetic energy per atom at 600 K
Now, substituting \( k = 1.38 \times 10^{-23} \, \text{J/K} \) and \( T = 600 \, \text{K} \):
\[ KE_{\text{atom}} = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 600 \]

### Step 5: Perform the calculation
Calculating this gives:
\[ KE_{\text{atom}} = \frac{3}{2} \times 1.38 \times 600 \times 10^{-23} \]
\[ KE_{\text{atom}} = \frac{3}{2} \times 828 \times 10^{-23} \]
\[ KE_{\text{atom}} = 1242 \times 10^{-23} \, \text{J} \]
\[ KE_{\text{atom}} = 1.24 \times 10^{-20} \, \text{J} \]

### Final Answer
The average thermal energy of a helium atom at 600 K is:
\[ \boxed{1.24 \times 10^{-20} \, \text{J}} \]

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