At `27^(@)C` and 2.0 atm pressure, the density of propene gas is :

Options

A. 10.1g`L^(−1)`
B. 3.41g`L^(−1)`
C. 1.1g`L^(−1)`
D. 9.1g`L^(−1)`

Explanation

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

### Step 1: Understand the Formula for Density
The density (D) of a gas can be calculated using the formula:
\[ D = \frac{PM}{RT} \]
where:
- \( P \) = pressure in atm
- \( M \) = molar mass of the gas in g/mol
- \( R \) = ideal gas constant (0.0821 atm·L/(K·mol))
- \( T \) = temperature in Kelvin

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

### Step 3: Identify the Values
Now, we need to identify the values for the variables:
- \( P = 2.0 \, atm \)
- \( M \) (molar mass of propene, C₃H₆) = 42 g/mol
- \( R = 0.0821 \, atm \cdot L/(K \cdot mol) \)
- \( T = 300 \, K \)

### Step 4: Substitute Values into the Density Formula
Now we can substitute the values into the density formula:
\[ D = \frac{PM}{RT} = \frac{(2.0 \, atm)(42 \, g/mol)}{(0.0821 \, atm \cdot L/(K \cdot mol))(300 \, K)} \]

### Step 5: Calculate the Density
Now, we will perform the calculations:
1. Calculate the numerator:
\[ 2.0 \, atm \times 42 \, g/mol = 84 \, g \cdot atm/mol \]

2. Calculate the denominator:
\[ 0.0821 \, atm \cdot L/(K \cdot mol) \times 300 \, K = 24.63 \, atm \cdot L/mol \]

3. Now divide the numerator by the denominator:
\[ D = \frac{84 \, g \cdot atm/mol}{24.63 \, atm \cdot L/mol} \approx 3.41 \, g/L \]

### Final Answer
The density of propene gas at 27°C and 2.0 atm pressure is approximately **3.41 g/L**.

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