The density of methane at 5.0 atm pressure and `27^0C` is :

Options

A. 1.9g`L^(−1)`
B. 2.8g`L^(−1)`
C. 3.2g`L^(−1)`
D. 0.32g`L^(−1)`

Explanation

To find the density of methane (CH₄) at a pressure of 5.0 atm and a temperature of 27°C, we can use the ideal gas law and the formula for density derived from it.

### Step-by-Step Solution:

1. **Identify the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

2. **Convert Temperature to Kelvin**:
The temperature given is 27°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Thus,
\[ T = 27 + 273 = 300 \, K \]

3. **Rearranging the Ideal Gas Law**:
We can express the number of moles \( n \) in terms of density \( D \):
\[ n = \frac{w}{M} \]
where \( w \) is the mass and \( M \) is the molar mass. The density \( D \) is defined as:
\[ D = \frac{w}{V} \]
By substituting \( n \) into the ideal gas law, we can rearrange it to find density:
\[ PV = \frac{w}{M}RT \implies D = \frac{PM}{RT} \]

4. **Substituting Known Values**:
- Pressure \( P = 5.0 \, atm \)
- Molar mass of methane \( M = 16 \, g/mol \)
- Gas constant \( R = 0.0821 \, \frac{atm \cdot L}{K \cdot mol} \)
- Temperature \( T = 300 \, K \)

Now, substituting these values into the density formula:
\[ D = \frac{(5.0 \, atm)(16 \, g/mol)}{(0.0821 \, \frac{atm \cdot L}{K \cdot mol})(300 \, K)} \]

5. **Calculating the Density**:
\[ D = \frac{80 \, g \cdot atm/mol}{24.63 \, atm \cdot L/mol} \approx 3.25 \, g/L \]

Rounding to two decimal places, we find:
\[ D \approx 3.2 \, g/L \]

### Final Answer:
The density of methane at 5.0 atm pressure and 27°C is approximately **3.2 g/L**.

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