A gas is filled in a vessel at `7^@ C`. If x fractional part escapes out at `27^@ C`. Find x. (Assuming pressure constant).

Answer

To solve the problem, we can use the principles of the kinetic theory of gases and the relationship between volume and temperature under constant pressure. Here’s a step-by-step solution:

### Step 1: Convert temperatures to Kelvin
The initial temperature \( T_1 \) is given as \( 7^\circ C \) and the final temperature \( T_2 \) is \( 27^\circ C \). We need to convert these temperatures to Kelvin.

\[ T_1 = 7 + 273 = 280 \, K \]
\[ T_2 = 27 + 273 = 300 \, K \]

### Step 2: Use the relationship between volume and temperature
Since the pressure is constant, we can use the relationship that states:

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

Let \( V_1 \) be the initial volume of the gas and \( V_2 \) be the final volume after some gas has escaped.

### Step 3: Express \( V_2 \) in terms of \( V_1 \)
Rearranging the equation gives us:

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

Substituting the values of \( T_1 \) and \( T_2 \):

\[ V_2 = V_1 \cdot \frac{300}{280} = V_1 \cdot \frac{15}{14} \]

### Step 4: Determine the volume that has escaped
If \( x \) is the fractional part that has escaped, then the volume that remains is:

\[ V_2 = V_1 - x V_1 \]

This can be rewritten as:

\[ V_2 = V_1(1 - x) \]

### Step 5: Set the two expressions for \( V_2 \) equal
Now we can set the two expressions for \( V_2 \) equal to each other:

\[ V_1(1 - x) = V_1 \cdot \frac{15}{14} \]

### Step 6: Simplify and solve for \( x \)
Dividing both sides by \( V_1 \) (assuming \( V_1 \neq 0 \)):

\[ 1 - x = \frac{15}{14} \]

Now, solving for \( x \):

\[ -x = \frac{15}{14} - 1 \]
\[ -x = \frac{15}{14} - \frac{14}{14} \]
\[ -x = \frac{1}{14} \]
\[ x = -\frac{1}{14} \]

Since \( x \) represents a fractional part that has escaped, we take the positive value:

\[ x = \frac{1}{14} \]

### Final Answer
The fractional part of the gas that has escaped is:

\[ x = \frac{1}{14} \]