A certain mass of gas at 273 K is expanded to 81 times its volume under adiabatic condition. If `lambda=1.25` for gas , then its final temperature is

Options

A. `-235^(@)C`
B. `-182^(@)C`
C. `-91^(@)C`
D. `-0^(@)C`

Answer

To solve the problem, we will use the adiabatic process equation for an ideal gas. The relationship we will use is:

\[ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \]

Where:
- \( T_1 \) is the initial temperature,
- \( T_2 \) is the final temperature,
- \( V_1 \) is the initial volume,
- \( V_2 \) is the final volume,
- \( \gamma \) (gamma) is the heat capacity ratio.

### Step 1: Identify the given values
- Initial temperature \( T_1 = 273 \, K \)
- Final volume \( V_2 = 81 \times V_1 \)
- Therefore, \( \frac{V_1}{V_2} = \frac{V_1}{81 V_1} = \frac{1}{81} \)
- \( \gamma = 1.25 \)

### Step 2: Substitute the values into the formula
We can substitute the known values into the equation:

\[ \frac{T_2}{273} = \left(\frac{1}{81}\right)^{1.25 - 1} \]

### Step 3: Simplify the exponent
Calculate \( \gamma - 1 \):

\[ \gamma - 1 = 1.25 - 1 = 0.25 \]

Now, we can rewrite the equation:

\[ \frac{T_2}{273} = \left(\frac{1}{81}\right)^{0.25} \]

### Step 4: Calculate \( \left(\frac{1}{81}\right)^{0.25} \)
We know that \( 81 = 3^4 \), thus:

\[ \left(\frac{1}{81}\right)^{0.25} = \left(3^{-4}\right)^{0.25} = 3^{-1} = \frac{1}{3} \]

### Step 5: Substitute back to find \( T_2 \)
Now we have:

\[ \frac{T_2}{273} = \frac{1}{3} \]

Multiplying both sides by 273 gives:

\[ T_2 = 273 \times \frac{1}{3} = 91 \, K \]

### Step 6: Final answer
The final temperature \( T_2 \) is:

\[ T_2 = 91 \, K \]

### Summary
The final temperature of the gas after expansion under adiabatic conditions is \( 91 \, K \).

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