2 gm of `O_(2)` at `27^(@)C` and 760 mm of Hg pressure has volume `"____________"`.

Options

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

Answer

To find the volume of 2 grams of \( O_2 \) at \( 27^\circ C \) and 760 mm of Hg pressure, we can use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure in atm
- \( V \) = volume in liters
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

### Step 1: Calculate the number of moles of \( O_2 \)
The molecular mass of \( O_2 \) is 32 g/mol. We can calculate the number of moles using the formula:

\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \, \text{g}}{32 \, \text{g/mol}} = \frac{1}{16} \, \text{mol} \]

### Step 2: Convert the temperature to Kelvin
The temperature in Celsius is given as \( 27^\circ C \). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 = 27 + 273 = 300 \, K \]

### Step 3: Convert the pressure to atm
The pressure is given as 760 mm of Hg. We know that:

\[ 1 \, \text{atm} = 760 \, \text{mm of Hg} \]

Thus, the pressure in atm is:

\[ P = 1 \, \text{atm} \]

### Step 4: Substitute the values into the Ideal Gas Law
Now we can substitute \( P \), \( n \), \( R \), and \( T \) into the Ideal Gas Law equation:

\[ PV = nRT \]

Substituting the values:

\[ 1 \, \text{atm} \cdot V = \left(\frac{1}{16} \, \text{mol}\right) \cdot (0.0821 \, \text{L·atm/(K·mol)}) \cdot (300 \, K) \]

### Step 5: Solve for \( V \)
Now we can solve for \( V \):

\[ V = \frac{\left(\frac{1}{16}\right) \cdot (0.0821) \cdot (300)}{1} \]

Calculating the right-hand side:

\[ V = \frac{0.0821 \cdot 300}{16} = \frac{24.63}{16} \approx 1.539375 \, \text{L} \]

Rounding to two decimal places, we get:

\[ V \approx 1.5 \, \text{L} \]

### Final Answer:
The volume of 2 grams of \( O_2 \) at \( 27^\circ C \) and 760 mm of Hg pressure is approximately **1.5 liters**.

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