A sample of gas occupies 100 mL al `27^@C` and 740 mm pressure. When the volume is changed to 80 mL at 740 mm pressure, the temperature of the gas will be

Options

A. `21.6^@C`
B. `240^@C`
C. `-33^@C`
D. `89.6^@C`

Explanation

To solve the problem, we will use Charles' Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature (in Kelvin). The formula can be expressed as:

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

### Step 1: Identify the given values
- Initial volume, \( V_1 = 100 \, \text{mL} \)
- Initial temperature, \( T_1 = 27^\circ C \)
- Final volume, \( V_2 = 80 \, \text{mL} \)
- Pressure remains constant at \( 740 \, \text{mm} \)

### Step 2: Convert the initial temperature from Celsius to Kelvin
To convert Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

So,

\[ T_1 = 27 + 273 = 300 \, K \]

### Step 3: Use Charles' Law to find the final temperature \( T_2 \)
We can rearrange Charles' Law to solve for \( T_2 \):

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

Substituting the known values:

\[ T_2 = \frac{80 \, \text{mL} \cdot 300 \, K}{100 \, \text{mL}} \]

### Step 4: Calculate \( T_2 \)
Now, we perform the calculation:

\[ T_2 = \frac{80 \cdot 300}{100} = \frac{24000}{100} = 240 \, K \]

### Step 5: Convert \( T_2 \) back to Celsius
To convert Kelvin back to Celsius, we use the formula:

\[ T(°C) = T(K) - 273 \]

So,

\[ T_2(°C) = 240 - 273 = -33^\circ C \]

### Final Answer
The temperature of the gas when the volume is changed to 80 mL at 740 mm pressure is:

\[ \boxed{-33^\circ C} \]

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