A gas in a cylinder. Its temperature is increased by `20%` on kelvin sacle and volume is reduced to `90%` how much percentage of the gas has to leak for pressure to remain constant?

Options

A. `20%`
B. `25%`
C. `30%`
D. `40%`

Explanation

To solve the problem step by step, we will use the ideal gas law and the information given in the question.

### Step 1: Understand the initial conditions
Let the initial temperature of the gas be \( T \) (in Kelvin) and the initial volume be \( V \). The initial number of moles of gas is \( n \).

### Step 2: Calculate the final temperature
The temperature is increased by \( 20\% \). Therefore, the final temperature \( T_f \) can be calculated as:
\[ T_f = T + 0.2T = 1.2T \]
However, since we need to express this in terms of the initial temperature, we can also express it as:
\[ T_f = \frac{6}{5} T \]

### Step 3: Calculate the final volume
The volume is reduced to \( 90\% \) of its original value. Therefore, the final volume \( V_f \) is:
\[ V_f = 0.9V = \frac{9}{10}V \]

### Step 4: Apply the ideal gas law before and after the changes
Before the changes, the ideal gas law is:
\[ PV = nRT \quad \text{(Equation 1)} \]
After the changes, the ideal gas law becomes:
\[ P \left(\frac{9}{10}V\right) = n_1 R \left(\frac{6}{5}T\right) \quad \text{(Equation 2)} \]
Where \( n_1 \) is the number of moles of gas remaining after some gas has leaked.

### Step 5: Set up the equation for constant pressure
Since the pressure \( P \) remains constant, we can equate the two equations:
\[ P \cdot \frac{9}{10}V = n_1 R \cdot \frac{6}{5}T \]
Dividing both sides by \( PV \) gives:
\[ \frac{9}{10} = \frac{n_1 \cdot \frac{6}{5}T}{nRT} \]
This simplifies to:
\[ \frac{9}{10} = \frac{n_1}{n} \cdot \frac{6}{5} \]

### Step 6: Solve for the ratio of \( n_1 \) to \( n \)
Rearranging the equation:
\[ \frac{n_1}{n} = \frac{9}{10} \cdot \frac{5}{6} \]
Calculating the right side:
\[ \frac{n_1}{n} = \frac{9 \cdot 5}{10 \cdot 6} = \frac{45}{60} = \frac{3}{4} \]

### Step 7: Calculate the percentage of gas that has leaked
If \( \frac{n_1}{n} = \frac{3}{4} \), then the fraction of gas that has leaked is:
\[ \text{Leaked gas} = n - n_1 = n - \frac{3}{4}n = \frac{1}{4}n \]
To find the percentage of gas leaked:
\[ \text{Percentage leaked} = \left(\frac{\frac{1}{4}n}{n}\right) \times 100 = 25\% \]

### Conclusion
Thus, the percentage of the gas that has to leak for the pressure to remain constant is **25%**.

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