An iar tight container having a lid with negli-gible mass and an area of `8cm^(2)` is partially evacuated. If a 48 N forces is required to pull the lig of the container and the atmospheric pressure is `1.0xx10^(5)Pa` the pressure in the container before it is opened must be

Options

A. 0.6 atm
B. 0.5 atm
C. 0.4 atm
D. 0.2 atm

Explanation

To solve the problem, we need to determine the pressure inside an airtight container before it is opened. The information provided includes the force required to pull the lid off the container and the atmospheric pressure. Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Lid
The lid of the container experiences two forces:
- The force due to the internal pressure (P_internal) acting upwards.
- The atmospheric pressure (P0) acting downwards, along with the additional force required to open the lid (48 N).

### Step 2: Write the Equation for Forces
From the balance of forces, we can write:
\[ P_{internal} \times A + F = P_0 \times A \]
Where:
- \( P_{internal} \) = internal pressure in the container (Pa)
- \( A \) = area of the lid (m²)
- \( F \) = force required to pull the lid (N)
- \( P_0 \) = atmospheric pressure (Pa)

### Step 3: Rearranging the Equation
Rearranging the equation to solve for \( P_{internal} \):
\[ P_{internal} = \frac{P_0 \times A - F}{A} \]

### Step 4: Convert Units
- Convert the area from cm² to m²:
\[ A = 8 \, \text{cm}^2 = 8 \times 10^{-4} \, \text{m}^2 \]
- Atmospheric pressure \( P_0 = 1.0 \times 10^5 \, \text{Pa} \)
- Force \( F = 48 \, \text{N} \)

### Step 5: Substitute Values into the Equation
Now substitute the values into the equation:
\[ P_{internal} = \frac{(1.0 \times 10^5 \, \text{Pa}) \times (8 \times 10^{-4} \, \text{m}^2) - 48 \, \text{N}}{8 \times 10^{-4} \, \text{m}^2} \]

### Step 6: Calculate the Internal Pressure
Calculate the numerator:
\[ P_{internal} = \frac{(1.0 \times 10^5 \times 8 \times 10^{-4}) - 48}{8 \times 10^{-4}} \]
\[ = \frac{80 - 48}{8 \times 10^{-4}} \]
\[ = \frac{32}{8 \times 10^{-4}} \]
\[ = 4 \times 10^5 \, \text{Pa} \]

### Step 7: Convert Pressure to Atmospheres
To convert the pressure from Pascals to atmospheres:
\[ P_{internal} = \frac{4 \times 10^5 \, \text{Pa}}{1.0 \times 10^5 \, \text{Pa/atm}} = 0.4 \, \text{atm} \]

### Final Answer
The pressure in the container before it is opened must be **0.4 atm**.

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