The sepific heat at constant volume for the monatomic argon is `0.075 kcal//kg-K`, whereas its gram molecular specific heat is `C_(V) = 2.98 cal//mol//K`. The mass of the argon atom is (Avogadro's number `= 6.02 xx 10^(23) "molecules"//"mol")

Options

A. `6.60 xx 10^(-23) g`
B. `3.30 xx 10^(-23) g`
C. `2.20 xx 10^(-23) g`
D. `13.20 xx 10^(-23) g`

Explanation

To find the mass of an argon atom, we will use the given specific heat at constant volume and the gram molecular specific heat. Here’s a step-by-step solution:

### Step 1: Understand the relationship between specific heat and molecular mass
The gram molecular specific heat at constant volume \( C_V \) is related to the molecular mass \( M \) of the gas and the specific heat at constant volume per unit mass \( C_{V}' \) as follows:

\[ C_V = M \cdot C_{V}' \]

Where:
- \( C_V \) is the gram molecular specific heat (in cal/mol·K)
- \( M \) is the molecular mass (in g/mol)
- \( C_{V}' \) is the specific heat at constant volume per unit mass (in cal/kg·K)

### Step 2: Convert the specific heat at constant volume from kcal to cal
Given:
- \( C_{V}' = 0.075 \) kcal/kg·K

To convert kcal to cal, we multiply by 1000:

\[ C_{V}' = 0.075 \times 1000 = 75 \text{ cal/kg·K} \]

### Step 3: Substitute the values into the equation
We know:
- \( C_V = 2.98 \) cal/mol·K
- \( C_{V}' = 75 \) cal/kg·K

Substituting these values into the equation:

\[ 2.98 = M \cdot 75 \]

### Step 4: Solve for the molecular mass \( M \)
Rearranging the equation to solve for \( M \):

\[ M = \frac{2.98}{75} \]

Calculating \( M \):

\[ M = \frac{2.98}{75} \approx 0.03973 \text{ kg/mol} = 39.73 \text{ g/mol} \]

### Step 5: Calculate the mass of one argon atom
To find the mass of a single argon atom, we use Avogadro's number (\( N_A = 6.02 \times 10^{23} \) molecules/mol):

\[ \text{Mass of one atom} = \frac{M}{N_A} \]

Substituting the values:

\[ \text{Mass of one atom} = \frac{39.73 \text{ g/mol}}{6.02 \times 10^{23} \text{ molecules/mol}} \]

Calculating the mass:

\[ \text{Mass of one atom} \approx \frac{39.73}{6.02 \times 10^{23}} \approx 6.60 \times 10^{-23} \text{ g} \]

### Final Answer
The mass of one argon atom is approximately \( 6.60 \times 10^{-23} \) g.

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