A conducting and closed container of capacity 100 liter contains an ideal gas at a high pressure. Now using a pump, the gas is taken out at a constant rate of 5 liter/sec. Find the time taken in which the pressure will decrease to initial `(P_("initial"))/(100)` ? (Assume isothermal condition)

Options

A. 46 sec
B. 92 sec
C. 118 sec
D. 146 sec

Explanation

To solve the problem step by step, we will use the ideal gas law and the concept of isothermal processes.

### Step 1: Understand the Problem
We have a closed container with a capacity of 100 liters filled with an ideal gas at high pressure. The gas is being pumped out at a constant rate of 5 liters per second. We need to find the time taken for the pressure to decrease to \( \frac{P_{\text{initial}}}{100} \).

### Step 2: Set Up the Variables
- Let \( V_0 = 100 \) liters (initial volume).
- Let \( P_{\text{initial}} \) be the initial pressure.
- The rate of gas removal is \( \frac{dV}{dt} = -5 \) liters/second (negative because the volume is decreasing).
- The final pressure we want to reach is \( P_{\text{final}} = \frac{P_{\text{initial}}}{100} \).

### Step 3: Relate Pressure and Volume
Using the ideal gas law:
\[ PV = nRT \]
For isothermal conditions, the product \( nRT \) remains constant. As gas is removed, the number of moles \( n \) decreases, leading to a decrease in pressure \( P \).

### Step 4: Express Change in Pressure
The change in pressure can be expressed as:
\[ dP = -\frac{P}{V_0} \cdot dV \]
Substituting \( dV = -5 dt \):
\[ dP = -\frac{P}{V_0} \cdot (-5 dt) = \frac{5P}{V_0} dt \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \frac{dP}{P} = \frac{5}{V_0} dt \]
Since \( V_0 = 100 \) liters:
\[ \frac{dP}{P} = \frac{5}{100} dt = \frac{1}{20} dt \]

### Step 6: Integrate the Equation
Integrate both sides:
\[ \int \frac{dP}{P} = \int \frac{1}{20} dt \]
This gives:
\[ \ln P = \frac{1}{20} t + C \]

### Step 7: Apply Limits
We need to find the time when \( P \) decreases from \( P_{\text{initial}} \) to \( \frac{P_{\text{initial}}}{100} \):
1. When \( t = 0 \), \( P = P_{\text{initial}} \).
2. When \( t = T \), \( P = \frac{P_{\text{initial}}}{100} \).

Substituting these limits into the integrated equation:
\[ \ln \left(\frac{P_{\text{initial}}}{100}\right) - \ln P_{\text{initial}} = \frac{1}{20} T \]
This simplifies to:
\[ \ln \left(\frac{1}{100}\right) = \frac{1}{20} T \]

### Step 8: Solve for Time \( T \)
The left side simplifies to:
\[ \ln \left(10^{-2}\right) = -2 \ln 10 \]
Thus:
\[ -2 \ln 10 = \frac{1}{20} T \]
Multiplying both sides by 20 gives:
\[ T = -40 \ln 10 \]

### Step 9: Calculate the Final Time
Using the approximate value \( \ln 10 \approx 2.303 \):
\[ T \approx -40 \times 2.303 \approx 92.12 \text{ seconds} \]

### Final Answer
The time taken for the pressure to decrease to \( \frac{P_{\text{initial}}}{100} \) is approximately **92 seconds**.

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