2gm of `O_(2)` at `27^(@)C` and 76mm of Hg pressure has volume

Options

A. 1.5 lit
B. 2.8 lit
C. 11.2 lit
D. 22.4 lit.

Explanation

To find the volume of 2 grams of \( O_2 \) at \( 27^\circ C \) and 76 mm of Hg pressure, we can use the Ideal Gas Law equation:

\[ PV = nRT \]

### Step 1: Convert the mass of \( O_2 \) to moles
The molar mass of \( O_2 \) is 32 g/mol.

\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \text{ g}}{32 \text{ g/mol}} = 0.0625 \text{ mol} \]

**Hint:** Remember to use the molar mass of the gas to convert grams to moles.

### Step 2: Convert the pressure from mm Hg to atm
The pressure in mm Hg is given as 76 mm Hg. To convert this to atm:

\[ P = \frac{76 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.1 \text{ atm} \]

**Hint:** Use the conversion factor of 760 mm Hg = 1 atm to convert pressure.

### Step 3: Convert the temperature from Celsius to Kelvin
The temperature is given as \( 27^\circ C \). To convert this to Kelvin:

\[ T = 27 + 273 = 300 \text{ K} \]

**Hint:** Remember to add 273 to the Celsius temperature to convert it to Kelvin.

### Step 4: Use the Ideal Gas Law to find the volume
Now we can substitute the values into the Ideal Gas Law equation:

\[ PV = nRT \]

Rearranging for \( V \):

\[ V = \frac{nRT}{P} \]

Substituting the known values:

\[ V = \frac{(0.0625 \text{ mol}) \times (0.082 \text{ L atm/(mol K)}) \times (300 \text{ K})}{0.1 \text{ atm}} \]

Calculating the numerator:

\[ 0.0625 \times 0.082 \times 300 = 1.5375 \]

Now, divide by the pressure:

\[ V = \frac{1.5375}{0.1} = 15.375 \text{ L} \]

### Final Answer
The volume of 2 grams of \( O_2 \) at \( 27^\circ C \) and 76 mm of Hg pressure is approximately **15.38 liters**.

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