A diesel engine takes in 1 mole of air at 300 K, 1 atm pressure and compresses it adiabatically to `(1)/(32)th` of the original volume. Considering air as a diatomic ideal gas, the change in temperature is

Options

A. 900 K
B. 1200 K
C. 600 K
D. 2400 K

Answer

To solve the problem step by step, we will use the principles of thermodynamics, specifically focusing on the adiabatic process for a diatomic ideal gas.

### Step 1: Understand the Adiabatic Process
In an adiabatic process, the following relation holds true for an ideal gas:
\[ PV^\gamma = \text{constant} \]
or
\[ TV^{\gamma - 1} = \text{constant} \]
where \( \gamma \) (gamma) is the adiabatic index, which for a diatomic gas (like air) is approximately \( \frac{7}{5} \).

### Step 2: Initial Conditions
Given:
- Initial temperature \( T_1 = 300 \, \text{K} \)
- Initial volume \( V_1 = V \)
- Final volume \( V_2 = \frac{V}{32} \)

### Step 3: Apply the Adiabatic Relation
Using the relation \( TV^{\gamma - 1} = \text{constant} \):
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Substituting the known values:
\[ 300 \cdot V^{\gamma - 1} = T_2 \cdot \left(\frac{V}{32}\right)^{\gamma - 1} \]

### Step 4: Calculate \( \gamma - 1 \)
First, calculate \( \gamma - 1 \):
\[ \gamma = \frac{7}{5} \implies \gamma - 1 = \frac{7}{5} - 1 = \frac{2}{5} \]

### Step 5: Substitute \( \gamma - 1 \) into the Equation
Now substitute \( \gamma - 1 \) back into the equation:
\[ 300 \cdot V^{\frac{2}{5}} = T_2 \cdot \left(\frac{V}{32}\right)^{\frac{2}{5}} \]

### Step 6: Simplify the Equation
Rearranging gives:
\[ T_2 = 300 \cdot \frac{V^{\frac{2}{5}}}{\left(\frac{V}{32}\right)^{\frac{2}{5}}} \]
This simplifies to:
\[ T_2 = 300 \cdot \frac{V^{\frac{2}{5}}}{\frac{V^{\frac{2}{5}}}{32^{\frac{2}{5}}}} = 300 \cdot 32^{\frac{2}{5}} \]

### Step 7: Calculate \( 32^{\frac{2}{5}} \)
Since \( 32 = 2^5 \):
\[ 32^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^2 = 4 \]

### Step 8: Final Temperature Calculation
Now substitute back:
\[ T_2 = 300 \cdot 4 = 1200 \, \text{K} \]

### Step 9: Calculate Change in Temperature
The change in temperature \( \Delta T \) is given by:
\[ \Delta T = T_2 - T_1 = 1200 \, \text{K} - 300 \, \text{K} = 900 \, \text{K} \]

### Final Answer
The change in temperature is \( \Delta T = 900 \, \text{K} \).

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