The volume of `2.8g` of `CO` at `27^(@)C` and `0.821 atm` pressure is `(R=0.0821` lit. atm `mol^(-1)K^(-1))`

Options

A. 1.5 litre
B. 3 litre
C. 30 litre
D. 0.3 litre

Explanation

To solve the problem of finding the volume of 2.8 g of carbon monoxide (CO) at 27°C and 0.821 atm pressure, we can use the Ideal Gas Law, which is expressed as:

\[ PV = nRT \]

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

### Step-by-Step Solution:

**Step 1: Convert the temperature from Celsius to Kelvin.**
- The temperature in Celsius is given as 27°C.
- To convert to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
\[ T = 27 + 273 = 300 \, K \]

**Step 2: Calculate the number of moles of CO.**
- The molecular weight of carbon monoxide (CO) is approximately 28 g/mol.
- The number of moles (\( n \)) can be calculated using the formula:
\[ n = \frac{W}{M} \]
Where:
- \( W \) = mass of the gas (2.8 g)
- \( M \) = molar mass of CO (28 g/mol)
\[ n = \frac{2.8 \, g}{28 \, g/mol} = 0.1 \, mol \]

**Step 3: Substitute the values into the Ideal Gas Law equation.**
- Rearranging the Ideal Gas Law to solve for volume \( V \):
\[ V = \frac{nRT}{P} \]
- Now substitute the values:
- \( n = 0.1 \, mol \)
- \( R = 0.0821 \, L·atm/(mol·K) \)
- \( T = 300 \, K \)
- \( P = 0.821 \, atm \)

\[ V = \frac{(0.1 \, mol) \times (0.0821 \, L·atm/(mol·K)) \times (300 \, K)}{0.821 \, atm} \]

**Step 4: Calculate the volume.**
\[ V = \frac{(0.1) \times (0.0821) \times (300)}{0.821} \]
\[ V = \frac{2.463}{0.821} \]
\[ V \approx 3.00 \, L \]

### Final Answer:
The volume of 2.8 g of CO at 27°C and 0.821 atm pressure is approximately **3.00 liters**.

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