What is the density of `CO_(2)` at `27 .^(@)C` and 2.5 atm pressure ?

Options

A. `5.2 g L^(-1)`
B. `6.2 g L^(-1)`
C. `7.3 g L^(-1)`
D. `4.46 g L^(-1)`

Explanation

To find the density of carbon dioxide (CO₂) at a temperature of 27°C and a pressure of 2.5 atm, we can use the ideal gas law in the form that relates density to pressure, temperature, and molar mass. Here’s a step-by-step solution:

### Step 1: Convert Temperature to Kelvin
The temperature in Celsius needs to be converted to Kelvin using the formula:
\[ T(K) = T(°C) + 273.15 \]
Given \( T = 27°C \):
\[ T = 27 + 273.15 = 300.15 \, K \]

### Step 2: Identify Molar Mass of CO₂
The molar mass of carbon dioxide (CO₂) can be calculated as follows:
- Molar mass of Carbon (C) = 12 g/mol
- Molar mass of Oxygen (O) = 16 g/mol
- Therefore, for CO₂ (1 C + 2 O):
\[ \text{Molar mass of CO₂} = 12 + (2 \times 16) = 12 + 32 = 44 \, g/mol \]

### Step 3: Use the Density Formula from Ideal Gas Law
The density (d) can be calculated using the rearranged ideal gas law:
\[ d = \frac{PM}{RT} \]
Where:
- \( P \) = pressure in atm = 2.5 atm
- \( M \) = molar mass of CO₂ = 44 g/mol
- \( R \) = ideal gas constant = 0.0821 atm·L/(mol·K)
- \( T \) = temperature in Kelvin = 300.15 K

### Step 4: Substitute Values into the Density Formula
Now we can substitute the values into the density formula:
\[ d = \frac{(2.5 \, \text{atm}) \times (44 \, \text{g/mol})}{(0.0821 \, \text{atm·L/(mol·K)}) \times (300.15 \, K)} \]

### Step 5: Calculate the Denominator
First, calculate the denominator:
\[ 0.0821 \times 300.15 = 24.63 \, \text{atm·L/mol} \]

### Step 6: Calculate the Density
Now substitute back into the density formula:
\[ d = \frac{110 \, g}{24.63 \, L} \]
\[ d \approx 4.46 \, g/L \]

### Final Answer
The density of CO₂ at 27°C and 2.5 atm pressure is approximately:
\[ \boxed{4.46 \, \text{g/L}} \]

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