What is the density (in g L–1) of CO2 at 400 K and exerting a pressure of 0.0821 atm? (R = 0.0821 L atm mol–1 K–1)
Answer & explanation
Correct answer: option 2
The correct answer is option 2. 0.11
To find the density (\(d\)) of \(CO_2\) at 400 K and exerting a pressure of 0.0821 atm, we can use the ideal gas law equation:
\(PV = nRT\)
By rearranging the equation, we can solve for density (\(d\)):
\(d = \frac{\text{molar mass} \times P}{R \times T}\)
Given:
\(P = 0.0821 \, \text{atm}\)
\(R = 0.0821 \, \text{L atm mol}^{-1} \text{ K}^{-1}\)
\(T = 400 \, \text{K}\)
Molar mass of \(CO_2 = 44 g/mol\)
Substituting the values into the equation, we get:
\(d = \frac{44 \, \text{g/mol} \times 0.0821 \, \text{atm}}{0.0821 \, \text{L atm mol}^{-1} \text{ K}^{-1} \times 400 \, \text{K}}\)
Simplifying the equation, we find:
\(d \approx 0.1105 \, \text{g/L}\)
Therefore, the density of \(CO_2\) at 400 K and 0.0821 atm is approximately 0.1105 g/L. The correct answer is (2) 0.11.