A sample of air occupies 10 L at `127^(@)C` and 1 atm pressure. What volume of air will be expelled when it is cooled to `-23^(@)C` at the same pressure?

Explanation

To solve the problem, we will use the combined gas law, which relates the volume, temperature, and pressure of a gas. The formula we will use is:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Where:
- \( V_1 \) = initial volume of the gas
- \( T_1 \) = initial temperature in Kelvin
- \( V_2 \) = final volume of the gas
- \( T_2 \) = final temperature in Kelvin

### Step 1: Convert the temperatures from Celsius to Kelvin
- The initial temperature \( T_1 = 127^\circ C = 127 + 273 = 400 \, K \)
- The final temperature \( T_2 = -23^\circ C = -23 + 273 = 250 \, K \)

### Step 2: Use the combined gas law to find \( V_2 \)
We know:
- \( V_1 = 10 \, L \)
- \( T_1 = 400 \, K \)
- \( T_2 = 250 \, K \)

Substituting the values into the equation:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Rearranging to find \( V_2 \):

\[ V_2 = V_1 \times \frac{T_2}{T_1} \]

Substituting the known values:

\[ V_2 = 10 \, L \times \frac{250 \, K}{400 \, K} \]

Calculating \( V_2 \):

\[ V_2 = 10 \, L \times 0.625 = 6.25 \, L \]

### Step 3: Calculate the volume of air expelled
The volume of air expelled is the difference between the initial volume and the final volume:

\[ \text{Volume expelled} = V_1 - V_2 \]

Substituting the values:

\[ \text{Volume expelled} = 10 \, L - 6.25 \, L = 3.75 \, L \]

### Final Answer:
The volume of air that will be expelled when it is cooled to \(-23^\circ C\) at the same pressure is **3.75 L**.

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