A stationary cylinder of oxygent used in a hospital has the following characteristics at room temperature `300 K`, gauge pressure `1.38 xx 10^(7)` Pa. volume `16 L`. If the flow area, measured at atmospheric pressure, is constant at `2.4 L//min`, the cylinder will last for nearly

Options

A. `5 h`
B. `10 h`
C. `15 h`
D. `20 h`

Explanation

To solve the problem step by step, we will use the ideal gas law and the information provided about the cylinder of oxygen.

### Step 1: Convert the volume of the cylinder from liters to cubic meters.
Given:
- Volume \( V_1 = 16 \, \text{L} = 16 \times 10^{-3} \, \text{m}^3 \)

### Step 2: Identify the initial conditions.
- Initial temperature \( T = 300 \, \text{K} \)
- Initial pressure \( P_1 = 1.38 \times 10^7 \, \text{Pa} \)
- Final pressure \( P_2 = P_{\text{atm}} = 10^5 \, \text{Pa} \)

### Step 3: Set up the equation using the ideal gas law.
Since the process is isothermal (constant temperature), we can use the relationship:
\[ P_1 V_1 = P_2 V_2 \]
Where \( V_2 \) is the volume of gas that flows out of the cylinder over time \( t \).

### Step 4: Express \( V_2 \) in terms of flow rate and time.
Given the flow rate:
- Flow rate = \( 2.4 \, \text{L/min} = 2.4 \times 10^{-3} \, \text{m}^3/\text{min} \)

Thus,
\[ V_2 = \text{Flow rate} \times t = (2.4 \times 10^{-3}) \times t \]

### Step 5: Substitute \( V_2 \) into the ideal gas equation.
Substituting \( V_2 \) into the equation:
\[ P_1 V_1 = P_2 (2.4 \times 10^{-3} \times t) \]

### Step 6: Rearrange the equation to solve for \( t \).
Rearranging gives:
\[ t = \frac{P_1 V_1}{P_2 \times 2.4 \times 10^{-3}} \]

### Step 7: Plug in the values and calculate \( t \).
Substituting the known values:
\[ t = \frac{(1.38 \times 10^7) \times (16 \times 10^{-3})}{(10^5) \times (2.4 \times 10^{-3})} \]

Calculating:
\[ t = \frac{(1.38 \times 10^7) \times (16 \times 10^{-3})}{(10^5) \times (2.4 \times 10^{-3})} \]
\[ t = \frac{(1.38 \times 16) \times 10^4}{2.4} \]
\[ t = \frac{22.08 \times 10^4}{2.4} \]
\[ t = 9.2 \times 10^3 \text{ minutes} \]
\[ t \approx 920 \text{ minutes} \]

### Step 8: Convert minutes to hours.
\[ t \approx \frac{920}{60} \approx 15.33 \text{ hours} \]

### Conclusion
The cylinder will last for nearly **15 hours**.

---