A vessel contains a gas under a pressure of `5 xx 10^(5)pa`. If `3//5` of the mass of the gas is flown out, What will be the gas pressure if the temperature being maintained constant,

Options

A. `50 MPa`
B. `2 MPa`
C. `0.2 MPa`
D. `0.5 MPa`

Explanation

To solve the problem step by step, we will use the ideal gas law and the relationship between pressure and mass when temperature and volume are constant.

### Step 1: Understand the initial conditions
The initial pressure of the gas in the vessel is given as:
\[ P_1 = 5 \times 10^5 \, \text{Pa} \]

### Step 2: Determine the initial mass
Let the initial mass of the gas be \( m \).

### Step 3: Calculate the final mass after gas is flown out
Since \( \frac{3}{5} \) of the mass of the gas is flown out, the remaining mass \( m_2 \) can be calculated as:
\[ m_2 = m - \frac{3}{5}m = \frac{2}{5}m \]

### Step 4: Use the relationship between pressure and mass
According to the ideal gas law, if the temperature is constant and the volume of the vessel does not change, the pressure of the gas is directly proportional to its mass. Therefore, we can write:
\[ \frac{P_2}{P_1} = \frac{m_2}{m_1} \]

### Step 5: Substitute the known values
We know:
- \( P_1 = 5 \times 10^5 \, \text{Pa} \)
- \( m_2 = \frac{2}{5}m \)
- \( m_1 = m \)

Substituting these into the equation gives:
\[ \frac{P_2}{5 \times 10^5} = \frac{\frac{2}{5}m}{m} \]
This simplifies to:
\[ \frac{P_2}{5 \times 10^5} = \frac{2}{5} \]

### Step 6: Solve for \( P_2 \)
Now, we can solve for \( P_2 \):
\[ P_2 = 5 \times 10^5 \times \frac{2}{5} \]
\[ P_2 = 2 \times 10^5 \, \text{Pa} \]

### Step 7: Convert to MegaPascal
To convert \( P_2 \) from Pascal to MegaPascal:
\[ P_2 = 2 \times 10^5 \, \text{Pa} = 0.2 \, \text{MPa} \]

### Final Answer
The final pressure of the gas in the vessel after \( \frac{3}{5} \) of the mass has flown out is:
\[ P_2 = 0.2 \, \text{MPa} \]

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