A sample of gas at `35^(@)C` and `1 atm` pressure occupies a volume of `3.75` litres. At what temperature should the gas be keep if it si required should the gas be keep if it is required to reduce the volume to `3` litres a the same pressure:

Options

A. `-26.6^(@)C`
B. `0^(@)C`
C. `3.98^(@)C`
D. `28^(@)C`

Explanation

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

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

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

### Step 1: Convert the initial temperature from Celsius to Kelvin
The initial temperature \( T_1 \) is given as \( 35^\circ C \). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

So,

\[ T_1 = 35 + 273 = 308 \, K \]

### Step 2: Identify the given values
From the problem, we have:
- \( V_1 = 3.75 \, L \)
- \( T_1 = 308 \, K \)
- \( V_2 = 3 \, L \)
- Pressure remains constant at \( 1 \, atm \)

### Step 3: Rearrange Charles's Law to solve for \( T_2 \)
We need to find \( T_2 \). Rearranging the formula gives us:

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

### Step 4: Substitute the known values into the equation
Now we can substitute the known values into the equation:

\[ T_2 = \frac{3 \, L}{3.75 \, L} \times 308 \, K \]

### Step 5: Calculate \( T_2 \)
Calculating the right side:

\[ T_2 = \frac{3}{3.75} \times 308 \]

Calculating \( \frac{3}{3.75} \):

\[ \frac{3}{3.75} = 0.8 \]

Now multiply by \( 308 \):

\[ T_2 = 0.8 \times 308 = 246.4 \, K \]

### Step 6: Convert \( T_2 \) back to Celsius
To convert \( T_2 \) back to Celsius, we use the formula:

\[ T(°C) = T(K) - 273 \]

So,

\[ T_2(°C) = 246.4 - 273 = -26.6^\circ C \]

### Final Answer
The temperature at which the gas should be kept to reduce the volume to 3 liters at the same pressure is:

\[ \boxed{-26.6^\circ C} \]