An open vessel at `27^(@)C` is heated until 3/8th of the air in it has been expelled. Assuming that the volume remains constant, calculate the temperature at which the vessel was heated

Options

A. `307^(@)C`
B. `107^(@)C`
C. `480^(@)C`
D. `207^(@)C`

Answer

To solve the problem, we will use the ideal gas law and the relationship between the number of moles of gas and temperature. Here is the step-by-step solution:

### Step 1: Understand the Problem
We have an open vessel at an initial temperature of 27°C, which is equivalent to 300 K (since K = °C + 273). When heated, 3/8 of the air is expelled from the vessel, meaning that 5/8 of the original air remains.

### Step 2: Set Up the Ideal Gas Law
In an open vessel, the pressure remains constant. According to the ideal gas law:
\[ PV = nRT \]
Since the volume (V) and pressure (P) are constant, we can equate the initial and final states of the gas:
\[ n_1RT_1 = n_2RT_2 \]
Where:
- \( n_1 \) = initial number of moles
- \( n_2 \) = final number of moles
- \( T_1 \) = initial temperature (300 K)
- \( T_2 \) = final temperature (unknown)

### Step 3: Determine the Number of Moles
Let’s assume the initial number of moles \( n_1 = 1 \). Since 3/8 of the air has been expelled, the remaining moles \( n_2 \) can be calculated as:
\[ n_2 = n_1 - \frac{3}{8}n_1 = 1 - \frac{3}{8} = \frac{5}{8} \]

### Step 4: Substitute Values into the Equation
Now we can substitute the values into the equation:
\[ n_1T_1 = n_2T_2 \]
\[ 1 \cdot 300 = \frac{5}{8} \cdot T_2 \]

### Step 5: Solve for \( T_2 \)
Rearranging the equation to solve for \( T_2 \):
\[ T_2 = \frac{300 \cdot 8}{5} \]
\[ T_2 = \frac{2400}{5} = 480 \text{ K} \]

### Step 6: Convert \( T_2 \) to Celsius
To convert from Kelvin to Celsius:
\[ T_2 = 480 - 273 = 207 \text{ °C} \]

### Final Answer
The temperature at which the vessel was heated is **207°C**.

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