At `27^(@)C` and 4.0 atm pressure, the density of propene gas is :

Options

A. 10.1g`L^(−1)`
B. 5.1g`L^(−1)`
C. 6.8g`L^(−1)`
D. 1.1g`L^(−1)`

Explanation

To find the density of propene gas at 27°C and 4.0 atm pressure, we can use the formula derived from the ideal gas law. Here’s a step-by-step solution:

### Step 1: Understand the formula for density
The density (D) of a gas can be expressed using the ideal gas equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

We can express the number of moles (n) as:
\[ n = \frac{w}{M} \]
Where:
- \( w \) = mass of the gas
- \( M \) = molar mass of the gas

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

Rearranging this leads to:
\[ \frac{w}{V} = \frac{PM}{RT} \]
The term \( \frac{w}{V} \) is the density (D) of the gas:
\[ D = \frac{PM}{RT} \]

### Step 2: Convert temperature to Kelvin
The temperature given is 27°C. To convert this to Kelvin:
\[ T(K) = T(°C) + 273 \]
\[ T = 27 + 273 = 300 \, K \]

### Step 3: Identify the values needed
- Pressure (P) = 4.0 atm
- Molar mass of propene (C₃H₆) = 42 g/mol
- Gas constant (R) = 0.0821 L·atm/(K·mol)

### Step 4: Substitute the values into the density formula
Now we can substitute the values into the density formula:
\[ D = \frac{PM}{RT} \]
\[ D = \frac{(4.0 \, \text{atm})(42 \, \text{g/mol})}{(0.0821 \, \text{L·atm/(K·mol)})(300 \, K)} \]

### Step 5: Calculate the density
Calculating the numerator:
\[ 4.0 \times 42 = 168 \, \text{g·atm/mol} \]

Calculating the denominator:
\[ 0.0821 \times 300 = 24.63 \, \text{L·atm/(mol)} \]

Now, substituting these values back into the density equation:
\[ D = \frac{168}{24.63} \approx 6.82 \, \text{g/L} \]

### Step 6: Final result
Thus, the density of propene gas at 27°C and 4.0 atm pressure is approximately:
\[ D \approx 6.8 \, \text{g/L} \]