A rigid container of negligible heat capacity contains one mole of an ideal gas. The temperatures of the gas increases by `1^@C` if `3.0` cal of heat is added to it. The gas may be

Options

A. helium
B. argon
C. oxygen
D. carbon dioxide

Answer

To solve the problem, we need to determine the type of ideal gas in a rigid container based on the heat added and the temperature change. Here's a step-by-step solution:

### Step 1: Understand the Given Information
We have:
- A rigid container (negligible heat capacity)
- 1 mole of an ideal gas
- Temperature increase: \( \Delta T = 1^\circ C \)
- Heat added: \( Q = 3.0 \, \text{cal} \)

### Step 2: Convert Heat from Calories to Joules
We know that:
1 calorie = 4.18 joules. Therefore, we can convert the heat added:
\[ Q = 3.0 \, \text{cal} \times 4.18 \, \text{J/cal} = 12.54 \, \text{J} \]

### Step 3: Use the Formula for Heat at Constant Volume
For an ideal gas at constant volume, the heat added is related to the change in temperature and the molar heat capacity at constant volume (\( C_v \)):
\[ Q = n C_v \Delta T \]
Where:
- \( n = 1 \, \text{mole} \)
- \( \Delta T = 1^\circ C \)

Substituting the values:
\[ 12.54 \, \text{J} = 1 \, \text{mole} \times C_v \times 1 \]
Thus, we find:
\[ C_v = 12.54 \, \text{J/mol} \]

### Step 4: Relate Molar Heat Capacity to Degrees of Freedom
The molar heat capacity at constant volume for an ideal gas can also be expressed in terms of the degrees of freedom (\( f \)):
\[ C_v = \frac{f}{2} R \]
Where \( R \) is the universal gas constant (\( R = 8.314 \, \text{J/(mol K)} \)).

### Step 5: Solve for Degrees of Freedom
Substituting \( C_v \) into the equation:
\[ 12.54 = \frac{f}{2} \times 8.314 \]
Rearranging gives:
\[ f = \frac{12.54 \times 2}{8.314} \]
Calculating this:
\[ f \approx \frac{25.08}{8.314} \approx 3.01 \]
Since \( f \) must be a whole number, we round it to \( f = 3 \).

### Step 6: Identify the Type of Gas
For an ideal gas, the degrees of freedom correspond to the type of gas:
- Monatomic gases (like Helium and Argon) have \( f = 3 \).
- Diatomic gases have \( f = 5 \).
- Polyatomic gases have \( f \) greater than 5.

Since we found \( f = 3 \), the gas must be a monatomic gas.

### Conclusion
The gas in the rigid container is likely to be either Helium or Argon.

---