An ideal gas of one mole is kept in a rigid container of negligible heat capacity. If `25J` of heat is supplied the gas temperature raises by `2^(@)C`. Then the gas may be

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the System
We have an ideal gas in a rigid container with negligible heat capacity. This means that the volume of the gas does not change (rigid container) and that the container does not absorb heat.

### Step 2: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta U = Q - W \]
Where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat added to the system,
- \(W\) is the work done by the system.

Since the container is rigid, the work done \(W = 0\). Therefore, we can simplify the equation to:
\[ \Delta U = Q \]

### Step 3: Substitute Known Values
We know from the problem that:
- Heat supplied \(Q = 25 \, \text{J}\)
- The temperature change \(\Delta T = 2 \, \text{°C} = 2 \, \text{K}\) (since the size of the degree is the same in Celsius and Kelvin).

Thus, we have:
\[ \Delta U = 25 \, \text{J} \]

### Step 4: Relate Change in Internal Energy to Molar Heat Capacity
The change in internal energy for an ideal gas can also be expressed as:
\[ \Delta U = n C_v \Delta T \]
Where:
- \(n\) is the number of moles (given as 1 mole),
- \(C_v\) is the molar heat capacity at constant volume,
- \(\Delta T\) is the change in temperature.

Substituting the known values:
\[ 25 \, \text{J} = 1 \, \text{mol} \cdot C_v \cdot 2 \, \text{K} \]

### Step 5: Solve for \(C_v\)
Rearranging the equation gives:
\[ C_v = \frac{25 \, \text{J}}{2 \, \text{K}} = 12.5 \, \text{J/K} \]

### Step 6: Determine the Type of Gas
The molar heat capacities for different types of gases are:
- Monatomic gases (like Helium and Argon): \(C_v = \frac{3}{2} R\)
- Diatomic gases (like Oxygen): \(C_v = \frac{5}{2} R\)
- Polyatomic gases (like Carbon Dioxide): \(C_v\) is greater than for diatomic gases.

Using the universal gas constant \(R \approx 8.31 \, \text{J/(mol K)}\):
\[ C_v \text{ (monatomic)} = \frac{3}{2} R = \frac{3}{2} \times 8.31 \approx 12.46 \, \text{J/K} \]
\[ C_v \text{ (diatomic)} = \frac{5}{2} R = \frac{5}{2} \times 8.31 \approx 20.78 \, \text{J/K} \]

Since \(C_v \approx 12.5 \, \text{J/K}\) is close to the value for a monatomic gas, we conclude that the gas could be either Helium or Argon.

### Final Answer
The gas may be Helium or Argon.

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