The volume of 2.89 g of carbon monoxide at `27^(@)C` and 0.821 atm pressure is

Options

A. 0.03L
B. 3L
C. 0.3L
D. 1.5L

Answer

To find the volume of 2.89 g of carbon monoxide (CO) at a temperature of 27°C and a pressure of 0.821 atm, 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))
- \( T \) = absolute temperature (in Kelvin)

### Step 1: Convert the temperature from Celsius to Kelvin
To convert Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

For our case:

\[ T = 27 + 273 = 300 \, K \]

### Step 2: Calculate the number of moles of carbon monoxide
The number of moles \( n \) can be calculated using the formula:

\[ n = \frac{W}{M} \]

Where:
- \( W \) = mass of the gas (in grams)
- \( M \) = molar mass of the gas (in g/mol)

The molar mass of carbon monoxide (CO) is:

\[ M = 12 \, (C) + 16 \, (O) = 28 \, g/mol \]

Now, we can calculate the number of moles:

\[ n = \frac{2.89 \, g}{28 \, g/mol} = 0.1032142857 \, mol \]

### Step 3: Rearrange the Ideal Gas Law to solve for volume \( V \)
We can rearrange the Ideal Gas Law to solve for volume \( V \):

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

### Step 4: Substitute the known values into the equation
Now we can substitute the values we have:

- \( n = 0.1032142857 \, mol \)
- \( R = 0.0821 \, L·atm/(K·mol) \)
- \( T = 300 \, K \)
- \( P = 0.821 \, atm \)

So,

\[ V = \frac{0.1032142857 \, mol \times 0.0821 \, L·atm/(K·mol) \times 300 \, K}{0.821 \, atm} \]

### Step 5: Calculate the volume
Now we can perform the calculation:

\[ V = \frac{0.1032142857 \times 0.0821 \times 300}{0.821} \]

Calculating the numerator:

\[ 0.1032142857 \times 0.0821 \times 300 \approx 2.537 \]

Now dividing by the pressure:

\[ V \approx \frac{2.537}{0.821} \approx 3.09 \, L \]

### Step 6: Round the final answer
Upon rounding, we can state that the volume of 2.89 g of carbon monoxide at 27°C and 0.821 atm pressure is approximately:

**3 liters**