The density of methane at 6.0 atm pressure and `27^0C` is :

Options

A. 1.9g`L^(−1)`
B. 3.89g`L^(−1)`
C. 0.19g`L^(−1)`
D. 9g`L^(−1)`

Explanation

To find the density of methane at a pressure of 6.0 atm and a temperature of 27°C, we can use the ideal gas equation and the formula for density derived from it.

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin**:
- The temperature is given as 27°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
- Therefore:
\[ T = 27 + 273.15 = 300.15 \text{ K} \approx 300 \text{ K} \]

2. **Identify the Molar Mass of Methane**:
- The molar mass (M) of methane (CH₄) is approximately 16 g/mol.

3. **Use the Ideal Gas Equation**:
- The ideal gas equation is given by:
\[ PV = nRT \]
- Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 atm·L/(K·mol))
- \( T \) = temperature (in K)

4. **Rearranging for Density**:
- Density (\( D \)) can be expressed as:
\[ D = \frac{m}{V} = \frac{n \cdot M}{V} \]
- From the ideal gas equation, we can express \( n \) as:
\[ n = \frac{PV}{RT} \]
- Substituting \( n \) into the density equation gives:
\[ D = \frac{PM}{RT} \]

5. **Substituting the Values**:
- Now we can substitute the known values into the density formula:
- \( P = 6.0 \) atm
- \( M = 16 \) g/mol
- \( R = 0.0821 \) atm·L/(K·mol)
- \( T = 300 \) K
- Thus:
\[ D = \frac{6.0 \text{ atm} \times 16 \text{ g/mol}}{0.0821 \text{ atm·L/(K·mol)} \times 300 \text{ K}} \]

6. **Calculating the Density**:
- Performing the calculation:
\[ D = \frac{96}{24.63} \approx 3.89 \text{ g/L} \]

### Final Answer:
The density of methane at 6.0 atm pressure and 27°C is approximately **3.89 g/L**.