At `27^(@)C` and `3.0` atm pressure, the density of propene gas is :

Options

A. `10.1 gL^(-1)`
B. `4.03 gL^(-1)`
C. `5.12 gL^(-1)`
D. `0.506 gL^(-1)`

Explanation

To find the density of propene gas at 27°C and 3.0 atm pressure, we can use the ideal gas law and the relationship between density, mass, and volume. Here’s a step-by-step solution:

### Step 1: Convert the Temperature to Kelvin
The temperature given is 27°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
\[ T = 27 + 273 = 300 \, K \]

**Hint:** Remember to always convert Celsius to Kelvin when using gas laws.

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

### Step 3: Rearrange the Ideal Gas Law to Find Density
Density (\(d\)) is defined as mass (\(m\)) divided by volume (\(V\)):
\[ d = \frac{m}{V} \]
From the ideal gas law, we can express volume as:
\[ V = \frac{nRT}{P} \]
Substituting this into the density equation gives:
\[ d = \frac{m}{\frac{nRT}{P}} = \frac{mP}{nRT} \]

### Step 4: Assume 1 Mole of Propene
For simplicity, we can assume \(n = 1\) mole of propene. The molar mass of propene (C₃H₆) is approximately 42 g/mol.

### Step 5: Substitute the Known Values
Now we can substitute the known values into the density equation:
- \(P = 3.0 \, atm\)
- \(m = 42 \, g\) (molar mass of propene)
- \(R = 0.0821 \, L·atm/(K·mol)\)
- \(T = 300 \, K\)

So the density becomes:
\[ d = \frac{42 \, g \cdot 3.0 \, atm}{1 \cdot 0.0821 \, L·atm/(K·mol) \cdot 300 \, K} \]

### Step 6: Calculate the Density
Calculating the above expression:
\[ d = \frac{126 \, g \cdot atm}{24.63 \, L·atm/(K·mol)} \approx 5.12 \, g/L \]

### Conclusion
The density of propene gas at 27°C and 3.0 atm pressure is approximately **5.12 g/L**.

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