The density of methane at 3.0 atm pressure and `27^0C` is :

Options

A. 1.9g`L^(−1)`
B. 2.9g`L^(−1)`
C. 0.19g`L^(−1)`
D. 3.8g`L^(−1)`

Explanation

To find the density of methane (CH₄) at a pressure of 3.0 atm and a temperature of 27°C, we can use the ideal gas law equation and the formula for density.

### Step-by-Step Solution:

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

2. **Use the Ideal Gas Law:**
- The ideal gas law 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)

3. **Express Density in Terms of Pressure, Molar Mass, and Temperature:**
- Density (\(\rho\)) can be expressed as:
\[ \rho = \frac{mass}{volume} = \frac{w}{V} \]
- From the ideal gas equation, we can rearrange it to find density:
\[ PM = \rho RT \]
- Therefore, density can be expressed as:
\[ \rho = \frac{PM}{RT} \]

4. **Identify the Molar Mass of Methane:**
- The molar mass of methane (CH₄) is:
\[ M = 12 \text{ (C)} + 4 \times 1 \text{ (H)} = 16 \text{ g/mol} \]

5. **Substitute Values into the Density Formula:**
- Now, substitute the values into the density formula:
- \(P = 3.0 \text{ atm}\)
- \(M = 16 \text{ g/mol}\)
- \(R = 0.0821 \text{ atm·L/(K·mol)}\)
- \(T = 300 \text{ K}\)
- Thus,
\[ \rho = \frac{(3.0 \text{ atm}) \times (16 \text{ g/mol})}{(0.0821 \text{ atm·L/(K·mol)}) \times (300 \text{ K})} \]

6. **Calculate the Density:**
- Performing the calculation:
\[ \rho = \frac{48}{24.63} \approx 1.95 \text{ g/L} \]

7. **Final Result:**
- The density of methane at 3.0 atm and 27°C is approximately:
\[ \rho \approx 1.95 \text{ g/L} \]