A cylinder contains 12 litres of oxygen at `20^(@)C` and 15 atm pressure. The temperature of the gas is raised to `35^(@)C` and its volume increased to 17 litres. What is the final pressure of gas (in atm)?

Options

A. 9
B. 11
C. 15
D. 17

Explanation

To solve the problem, we will use the ideal gas law and the relationship between the initial and final states of the gas. The ideal gas law states that:

\[ PV = nRT \]

Since the number of moles (n) and the gas constant (R) remain constant, we can write the equation for the initial and final states of the gas as follows:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( T_1 \) = initial temperature (in Kelvin)
- \( P_2 \) = final pressure
- \( V_2 \) = final volume
- \( T_2 \) = final temperature (in Kelvin)

### Step 1: Convert temperatures from Celsius to Kelvin
- Initial temperature \( T_1 = 20^\circ C = 20 + 273 = 293 \, K \)
- Final temperature \( T_2 = 35^\circ C = 35 + 273 = 308 \, K \)

### Step 2: Identify the known values
- \( P_1 = 15 \, atm \)
- \( V_1 = 12 \, L \)
- \( T_1 = 293 \, K \)
- \( V_2 = 17 \, L \)
- \( T_2 = 308 \, K \)

### Step 3: Substitute the known values into the equation
Using the equation:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Substituting the known values:

\[ \frac{15 \, atm \times 12 \, L}{293 \, K} = \frac{P_2 \times 17 \, L}{308 \, K} \]

### Step 4: Solve for \( P_2 \)
Rearranging the equation to solve for \( P_2 \):

\[ P_2 = \frac{15 \, atm \times 12 \, L \times 308 \, K}{293 \, K \times 17 \, L} \]

### Step 5: Calculate \( P_2 \)
Now, we can perform the calculation:

1. Calculate the numerator:
\[ 15 \times 12 \times 308 = 56160 \]

2. Calculate the denominator:
\[ 293 \times 17 = 4981 \]

3. Now divide:
\[ P_2 = \frac{56160}{4981} \approx 11.29 \, atm \]

### Step 6: Round to appropriate significant figures
The final pressure \( P_2 \) is approximately \( 11 \, atm \).

### Final Answer:
The final pressure of the gas is approximately **11 atm**.

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