Density of methane , at `250^(@)C` and 6 atm pressure, is `[R=0.0821 atm ] :`

Options

A. `2.236 g//L`
B. `8 g//L`
C. `12 g//L`
D. `16 g //L`

Explanation

To find the density of methane at 250°C and 6 atm pressure, we can use the ideal gas law and the relationship between density, mass, and volume. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin:**
\[ T(K) = T(°C) + 273 \]
Given \( T = 250°C \):
\[ T = 250 + 273 = 523 \, K \]

2. **Identify Given Values:**
- Pressure \( P = 6 \, atm \)
- Gas constant \( R = 0.0821 \, \frac{L \cdot atm}{K \cdot mol} \)
- Molecular weight of methane \( M = 16 \, g/mol \)

3. **Use the Ideal Gas Equation:**
The ideal gas equation is:
\[ PV = nRT \]
where \( n \) is the number of moles. We can express \( n \) in terms of mass \( W \) and molecular weight \( M \):
\[ n = \frac{W}{M} \]
Substituting this into the ideal gas equation gives:
\[ PV = \frac{W}{M}RT \]

4. **Rearranging the Equation:**
Rearranging the equation to find the density \( \rho \) (where \( \rho = \frac{W}{V} \)):
\[ PV = \frac{W}{M}RT \implies W = \frac{PVM}{RT} \]
Dividing both sides by \( V \):
\[ \frac{W}{V} = \frac{PM}{RT} \]
Thus, the density \( \rho \) can be expressed as:
\[ \rho = \frac{PM}{RT} \]

5. **Substituting the Values:**
Now substitute the known values into the density equation:
\[ \rho = \frac{(6 \, atm)(16 \, g/mol)}{(0.0821 \, \frac{L \cdot atm}{K \cdot mol})(523 \, K)} \]

6. **Calculating the Density:**
First, calculate the numerator:
\[ 6 \times 16 = 96 \, g \cdot atm/mol \]
Now calculate the denominator:
\[ 0.0821 \times 523 \approx 42.94 \, L \cdot atm/mol \]
Now divide the two results:
\[ \rho = \frac{96}{42.94} \approx 2.236 \, g/L \]

7. **Final Result:**
The density of methane at 250°C and 6 atm pressure is approximately:
\[ \rho \approx 2.236 \, g/L \]