At 127°C and l atm pressure, a mixture of a gas contains 0.3 mole of `N_(2)`, 0.2 mole of `O_(2)` The volume of the mixture is

Options

A. 15 lit
B. 22.4 lit
C. 18.2 lit
D. 16.4 lit

Answer

To find the volume of a gas mixture containing 0.3 moles of nitrogen (N₂) and 0.2 moles of oxygen (O₂) at a temperature of 127°C and a pressure of 1 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 gas
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step 1: Convert Temperature to Kelvin
First, we need to convert the temperature from Celsius to Kelvin.

\[ T(K) = T(°C) + 273 \]

Given:
\[ T(°C) = 127 \]

Calculating:
\[ T(K) = 127 + 273 = 400 \, K \]

### Step 2: Calculate Total Moles of Gas
Next, we find the total number of moles of the gas mixture.

Given:
- Moles of \( N_2 = 0.3 \)
- Moles of \( O_2 = 0.2 \)

Calculating total moles \( n \):
\[ n = 0.3 + 0.2 = 0.5 \, \text{moles} \]

### Step 3: Use the Ideal Gas Law to Find Volume
Now we can substitute the values into the Ideal Gas Law equation to find the volume \( V \).

Given:
- \( P = 1 \, \text{atm} \)
- \( n = 0.5 \, \text{moles} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 400 \, K \)

Substituting into the equation:
\[ V = \frac{nRT}{P} \]

Calculating:
\[ V = \frac{0.5 \times 0.0821 \times 400}{1} \]

\[ V = 0.5 \times 0.0821 \times 400 = 16.42 \, \text{liters} \]

### Final Answer
The volume of the gas mixture is approximately:
\[ V \approx 16.4 \, \text{liters} \]

---