Nitrogen gas is filled in an isolated container. If `alpha` fraction of moles dissociates without exchange of any energy, then the fractional change in its temperature is

Options

A. a.`(- alpha )/(5 + alpha)`
B. b.`( alpha)/(3 + alpha)`
C. c.`(- 3 alpha)/(2 + alpha)`
D. d.`5 alpha/(2 + 3 alpha)`

Answer

To solve the problem of finding the fractional change in temperature when a fraction of nitrogen gas dissociates in an isolated container, we can follow these steps:

### Step 1: Understanding the Initial Conditions
We start with nitrogen gas (N₂) in an isolated container. Let the initial number of moles of nitrogen be \( n \) and the initial temperature be \( T_1 \).

### Step 2: Internal Energy Expression
The internal energy \( E \) of a diatomic gas can be expressed as:
\[ E = n C_v T \]
For nitrogen (a diatomic gas), the molar heat capacity at constant volume \( C_v \) is given by:
\[ C_v = \frac{5R}{2} \]
Thus, the initial internal energy can be written as:
\[ E_1 = n \cdot \frac{5R}{2} \cdot T_1 \]

### Step 3: Temperature in Terms of Internal Energy
From the expression for internal energy, we can rearrange it to find the initial temperature:
\[ T_1 = \frac{2E_1}{5nR} \]

### Step 4: Analyzing the Dissociation
If \( \alpha \) fraction of the moles dissociate, the number of moles of nitrogen remaining after dissociation is:
\[ n_1 = n(1 - \alpha) \]
The dissociation of nitrogen gas produces nitrogen atoms (N), which are considered as monoatomic gas. The number of moles of nitrogen atoms produced is:
\[ n_2 = n \cdot 2\alpha \]

### Step 5: Internal Energy After Dissociation
After dissociation, the internal energy can be expressed as:
\[ E_2 = n_1 C_{v, diatomic} T_2 + n_2 C_{v, monoatomic} T_2 \]
Where:
- \( C_{v, diatomic} = \frac{5R}{2} \)
- \( C_{v, monoatomic} = \frac{3R}{2} \)

Substituting the values:
\[ E_2 = n(1 - \alpha) \cdot \frac{5R}{2} \cdot T_2 + n(2\alpha) \cdot \frac{3R}{2} \cdot T_2 \]

### Step 6: Simplifying the Energy Equation
Combining the terms:
\[ E_2 = nT_2 \left( (1 - \alpha) \cdot \frac{5R}{2} + 2\alpha \cdot \frac{3R}{2} \right) \]
\[ E_2 = nT_2 \cdot \frac{R}{2} \left( 5 - 5\alpha + 6\alpha \right) \]
\[ E_2 = nT_2 \cdot \frac{R}{2} (5 + \alpha) \]

### Step 7: Setting Up the Energy Conservation
Since the container is isolated, the internal energy remains constant:
\[ E_1 = E_2 \]
Thus:
\[ n \cdot \frac{5R}{2} \cdot T_1 = nT_2 \cdot \frac{R}{2} (5 + \alpha) \]

### Step 8: Solving for \( T_2 \)
Canceling \( n \) and \( \frac{R}{2} \) from both sides:
\[ 5T_1 = T_2(5 + \alpha) \]
Solving for \( T_2 \):
\[ T_2 = \frac{5T_1}{5 + \alpha} \]

### Step 9: Finding the Fractional Change in Temperature
The fractional change in temperature is given by:
\[ \text{Fractional Change} = \frac{T_2 - T_1}{T_1} \]
Substituting \( T_2 \):
\[ \text{Fractional Change} = \frac{\frac{5T_1}{5 + \alpha} - T_1}{T_1} \]
\[ = \frac{5 - (5 + \alpha)}{5 + \alpha} \]
\[ = \frac{-\alpha}{5 + \alpha} \]

### Final Answer
Thus, the fractional change in temperature is:
\[ \text{Fractional Change} = -\frac{\alpha}{5 + \alpha} \]