A gas is filled in a cylinder. its temperature is increased by 20% on the Kelvin scale and volume is reduced by 10% How much percentage of the gas has to leak for pressure to remind constant ?

Options

A. `30%`
B. `40%`
C. `15%`
D. `25%`

Answer

To solve the problem, we will use the ideal gas law, which states:

\[ P = \frac{nRT}{V} \]

Where:
- \( P \) = Pressure
- \( n \) = Number of moles of gas
- \( R \) = Universal gas constant
- \( T \) = Temperature in Kelvin
- \( V \) = Volume

### Step 1: Define the initial conditions
Let:
- Initial number of moles of gas = \( n_1 \)
- Initial temperature = \( T_1 \)
- Initial volume = \( V_1 \)
- Initial pressure = \( P_1 \)

From the problem, we know:
- The temperature is increased by 20%, so:
\[ T_2 = T_1 + 0.2T_1 = 1.2T_1 \]
- The volume is reduced by 10%, so:
\[ V_2 = V_1 - 0.1V_1 = 0.9V_1 \]

### Step 2: Set up the equation for constant pressure
Since the pressure must remain constant (\( P_1 = P_2 \)), we can write the equations for pressure before and after the changes:

Initial pressure:
\[ P_1 = \frac{n_1RT_1}{V_1} \]

Final pressure:
\[ P_2 = \frac{n_2RT_2}{V_2} \]

Where \( n_2 \) is the number of moles after some gas has leaked.

### Step 3: Equate the pressures
Since \( P_1 = P_2 \):
\[ \frac{n_1RT_1}{V_1} = \frac{n_2R(1.2T_1)}{0.9V_1} \]

### Step 4: Simplify the equation
Cancel \( R \) and \( V_1 \) from both sides:
\[ n_1T_1 = n_2(1.2T_1/0.9) \]

### Step 5: Solve for \( n_2 \)
This simplifies to:
\[ n_2 = \frac{n_1 \cdot 0.9}{1.2} \]

### Step 6: Calculate the percentage of gas that has leaked
The amount of gas that has leaked is:
\[ \text{Leaked gas} = n_1 - n_2 = n_1 - \frac{n_1 \cdot 0.9}{1.2} \]

Calculating this gives:
\[ = n_1 \left(1 - \frac{0.9}{1.2}\right) \]
\[ = n_1 \left(1 - 0.75\right) \]
\[ = n_1 \cdot 0.25 \]

The percentage of gas that has leaked is:
\[ \text{Percentage leaked} = \frac{0.25n_1}{n_1} \times 100\% = 25\% \]

### Final Answer
Thus, **25% of the gas has to leak for the pressure to remain constant.**

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