The density of Carbon dioxide at 4.0 atm pressure and `27^0C` is :

Options

A. 8.9g`L^(−1)`
B. 7.9g`L^(−1)`
C. 7.14g`L^(−1)`
D. 8g`L^(−1)`

Explanation

To find the density of carbon dioxide (CO₂) at a pressure of 4.0 atm and a temperature of 27°C, we can use the ideal gas equation and the relationship between density and the ideal gas law. Here’s a step-by-step solution:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is expressed as:
\[ PV = nRT \]
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 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: Relate Moles to Density
The number of moles \( n \) can also be expressed in terms of mass \( w \) and molar mass \( M \):
\[ n = \frac{w}{M} \]
Substituting this into the ideal gas equation gives:
\[ PV = \frac{w}{M}RT \]
Rearranging this, we can express density \( \rho \) (where \( \rho = \frac{w}{V} \)):
\[ P = \frac{\rho M}{RT} \]
Thus, the density can be expressed as:
\[ \rho = \frac{PM}{RT} \]

### Step 4: Substitute Known Values
Now we can substitute the known values into the density formula:
- \( P = 4.0 \, \text{atm} \)
- \( M = 44 \, \text{g/mol} \) (molar mass of CO₂)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 300 \, K \)

Substituting these values into the density equation:
\[ \rho = \frac{(4.0 \, \text{atm}) \times (44 \, \text{g/mol})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (300 \, K)} \]

### Step 5: Calculate the Density
Now perform the calculation:
1. Calculate the numerator:
\[ 4.0 \times 44 = 176 \, \text{g·atm/mol} \]
2. Calculate the denominator:
\[ 0.0821 \times 300 = 24.63 \, \text{L·atm/mol} \]
3. Now divide the numerator by the denominator:
\[ \rho = \frac{176}{24.63} \approx 7.14 \, \text{g/L} \]

### Final Answer
The density of carbon dioxide at 4.0 atm and 27°C is approximately:
\[ \rho \approx 7.14 \, \text{g/L} \]