The density of methane at 2.0 atm pressure and `27^0C` is :

Options

A. 1.30g`L^(−1)`
B. 0.13g`L^(−1)`
C. 0.26g`L^(−1)`
D. 26.0g`L^(−1)`

Answer

To find the density of methane at a pressure of 2.0 atm and a temperature of 27°C, we can use the ideal gas law and the formula for density derived from it. Here’s the step-by-step solution:

### Step 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
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Relate Moles to Density
We can express the number of moles \( n \) as:
\[ n = \frac{W}{M} \]
where:
- \( W \) = weight of the gas
- \( M \) = molar mass of the gas

Substituting this into the ideal gas law gives:
\[ PV = \frac{W}{M}RT \]

### Step 3: Rearranging for Density
We can rearrange this equation to express density (\( \rho \)):
\[ P = \frac{W}{V} \cdot \frac{RT}{M} \]
Here, \( \frac{W}{V} \) is the density (\( \rho \)), so we can write:
\[ \rho = \frac{PM}{RT} \]

### Step 4: Identify Values
For methane (\( CH_4 \)):
- Molar mass \( M = 16 \, \text{g/mol} \)
- Pressure \( P = 2.0 \, \text{atm} \)
- Temperature \( T = 27^\circ C = 300 \, \text{K} \) (convert Celsius to Kelvin by adding 273)
- Universal gas constant \( R = 0.0821 \, \text{atm L K}^{-1} \text{mol}^{-1} \)

### Step 5: Substitute Values into the Density Formula
Now, substituting the values into the density formula:
\[ \rho = \frac{(2.0 \, \text{atm})(16 \, \text{g/mol})}{(0.0821 \, \text{atm L K}^{-1} \text{mol}^{-1})(300 \, \text{K})} \]

### Step 6: Calculate the Density
Calculating the denominator:
\[ 0.0821 \times 300 = 24.63 \, \text{atm L/mol} \]
Now substituting back into the density equation:
\[ \rho = \frac{32}{24.63} \approx 1.297 \, \text{g/L} \]

### Step 7: Final Result
Thus, the density of methane at 2.0 atm and 27°C is approximately:
\[ \rho \approx 1.30 \, \text{g/L} \]

### Summary
The density of methane at 2.0 atm and 27°C is approximately **1.30 g/L**.

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