If for an ideal gas with 2 litres volume, pressure was increased by 0.25 atm then volume became 555 ml. At what initial pressure was the gas present?

Options

A. 0.096 mm Hg
B. 0.96 mm Hg
C. 73 mm Hg
D. 73 atm

Answer

To solve the problem, we will use the ideal gas law and the relationship between pressure and volume. The steps are as follows:

### Step 1: Identify the known values
- Initial volume (V1) = 2 liters = 2000 ml
- Final volume (V2) = 555 ml
- Increase in pressure = 0.25 atm

### Step 2: Define the pressures
Let the initial pressure be P1 (in atm). Then the final pressure (P2) can be expressed as:
\[ P2 = P1 + 0.25 \]

### Step 3: Apply the ideal gas law (Boyle's Law)
According to Boyle's Law, for a given amount of gas at constant temperature:
\[ P1 \times V1 = P2 \times V2 \]

Substituting the known values:
\[ P1 \times 2000 = (P1 + 0.25) \times 555 \]

### Step 4: Expand and rearrange the equation
Expanding the equation:
\[ 2000 P1 = 555 P1 + 0.25 \times 555 \]

Rearranging gives:
\[ 2000 P1 - 555 P1 = 0.25 \times 555 \]

### Step 5: Combine like terms
Combine the terms involving P1:
\[ (2000 - 555) P1 = 0.25 \times 555 \]
\[ 1445 P1 = 0.25 \times 555 \]

### Step 6: Calculate the right side
Calculating the right side:
\[ 0.25 \times 555 = 138.75 \]

### Step 7: Solve for P1
Now we can solve for P1:
\[ P1 = \frac{138.75}{1445} \]

Calculating this gives:
\[ P1 \approx 0.096 atm \]

### Step 8: Convert atm to mmHg
To convert atm to mmHg, we use the conversion factor:
\[ 1 atm = 760 mmHg \]
So,
\[ P1 = 0.096 \times 760 \approx 73 mmHg \]

### Final Answer
The initial pressure of the gas was approximately **73 mmHg**.

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