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 will use the ideal gas law and the relationship between volume and temperature at constant pressure. Here’s a step-by-step solution:

### Step 1: Convert Temperatures to Kelvin
We need to convert the given temperatures from Celsius to Kelvin because the ideal gas law requires absolute temperatures.

- Initial temperature \( T_1 = 7^\circ C = 7 + 273 = 280 \, K \)
- Final temperature \( T_2 = 27^\circ C = 27 + 273 = 300 \, K \)

**Hint:** Remember that to convert Celsius to Kelvin, you add 273.

### Step 2: Use the Relationship Between Volume and Temperature
Since the pressure is constant, we can use the relationship that the volume of a gas is directly proportional to its temperature (in Kelvin):

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

Let’s denote the initial volume as \( V_1 = V \) and the final volume as \( V_2 \).

**Hint:** This relationship comes from the ideal gas law, where \( PV = nRT \).

### Step 3: Substitute the Known Values
Now we can substitute the values of \( T_1 \) and \( T_2 \) into the equation:

\[ \frac{V}{280} = \frac{V_2}{300} \]

### Step 4: Solve for \( V_2 \)
Cross-multiplying gives us:

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

**Hint:** Cross-multiplication is a useful technique to isolate variables in equations.

### Step 5: Calculate the Change in Volume
The change in volume \( \Delta V \) is given by:

\[ \Delta V = V_2 - V_1 = \left( \frac{15}{14} V \right) - V = \left( \frac{15}{14} V - \frac{14}{14} V \right) = \frac{1}{14} V \]

### Step 6: Find the Fractional Part Escaped
The fractional part of the gas that escapes is given by:

\[ x = \frac{\Delta V}{V_1} = \frac{\frac{1}{14} V}{V} = \frac{1}{14} \]

Thus, the fractional part \( x \) that escapes is:

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

### Final Answer
The value of \( x \) is \( \frac{1}{14} \).

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