`40` g of Argon is heated from `40^(@)C` to `100^(@)C (R=2 cal//mol.)` What is the heat absorbed at the constant volume by the Argon?

Explanation

To solve the problem of calculating the heat absorbed by 40 g of Argon when heated from 40°C to 100°C at constant volume, we can follow these steps:

### Step 1: Identify the properties of Argon
Argon (Ar) is a monoatomic gas. The molar mass of Argon is approximately 40 g/mol.

### Step 2: Calculate the number of moles of Argon
To find the number of moles (N) of Argon, we use the formula:
\[ N = \frac{\text{mass}}{\text{molar mass}} \]
Given:
- Mass of Argon = 40 g
- Molar mass of Argon = 40 g/mol

So,
\[ N = \frac{40 \, \text{g}}{40 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 3: Determine the heat capacity at constant volume (Cv)
For a monoatomic gas, the heat capacity at constant volume (Cv) is given by:
\[ C_v = \frac{3}{2} R \]
Where R is the universal gas constant. Given R = 2 cal/mol·K, we can calculate Cv:
\[ C_v = \frac{3}{2} \times 2 \, \text{cal/mol·K} = 3 \, \text{cal/mol·K} \]

### Step 4: Calculate the change in temperature (ΔT)
The change in temperature (ΔT) is calculated as:
\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \]
Given:
- Initial temperature (T_initial) = 40°C
- Final temperature (T_final) = 100°C

So,
\[ \Delta T = 100°C - 40°C = 60°C \]

### Step 5: Calculate the heat absorbed (Q)
The heat absorbed at constant volume can be calculated using the formula:
\[ Q = N \cdot C_v \cdot \Delta T \]
Substituting the values we found:
- N = 1 mol
- \( C_v = 3 \, \text{cal/mol·K} \)
- \( \Delta T = 60 \, \text{K} \)

Thus,
\[ Q = 1 \, \text{mol} \cdot 3 \, \text{cal/mol·K} \cdot 60 \, \text{K} = 180 \, \text{cal} \]

### Final Answer
The heat absorbed by the Argon gas is **180 calories**.

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