Find the density of methane at 298K and 72cm of Hg pressure 

Answer

To find the density of methane (CH₄) at a temperature of 298 K and a pressure of 72 cm of Hg, we can follow these steps:

### Step 1: Convert Pressure from cm of Hg to Atmospheres
The pressure given is 72 cm of Hg. We need to convert this to atmospheres.

1 atm = 760 mm Hg

To convert 72 cm of Hg to mm Hg:
\[ 72 \, \text{cm} = 720 \, \text{mm Hg} \]

Now, convert mm Hg to atm:
\[ P = \frac{720 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} \approx 0.947 \, \text{atm} \]

### Step 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
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

### Step 3: Relate Moles to Mass and Molecular Weight
The number of moles \( n \) can be expressed as:
\[ n = \frac{m}{M} \]
Where:
- \( m \) = mass in grams
- \( M \) = molecular weight of methane (CH₄), which is 16 g/mol.

### Step 4: Substitute into the Ideal Gas Law
Substituting \( n \) into the ideal gas law gives:
\[ PV = \frac{m}{M}RT \]

Rearranging for mass:
\[ m = \frac{PVM}{RT} \]

### Step 5: Calculate Density
Density \( D \) is defined as:
\[ D = \frac{m}{V} \]
Substituting for \( m \):
\[ D = \frac{P \cdot M}{R \cdot T} \]

### Step 6: Substitute Values
Now, we can substitute the known values:
- \( P = 0.947 \, \text{atm} \)
- \( M = 16 \, \text{g/mol} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 298 \, \text{K} \)

So:
\[ D = \frac{0.947 \, \text{atm} \cdot 16 \, \text{g/mol}}{0.0821 \, \text{L·atm/(K·mol)} \cdot 298 \, \text{K}} \]

### Step 7: Calculate the Final Density
Calculating the above expression:
\[ D = \frac{15.152}{24.4758} \approx 0.618 \, \text{g/L} \]

### Conclusion
The density of methane at 298 K and 72 cm of Hg pressure is approximately **0.618 g/L**.

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