Question

At $300\ \mathrm{K}$, $22\ \mathrm{g}$ of $\mathrm{CO_2}$ gas exerts a pressure of $5\ \mathrm{atmosphere}$. What is the volume of the gas at the same temperature? ($R = 0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$)

• A. $5.61\ \mathrm{dm^3}$
• B. $8.20\ \mathrm{dm^3}$
• C. $2.46\ \mathrm{dm^3}$
• D. $3.80\ \mathrm{dm^3}$

Hint:

To streamline mental math with the gas constant, group terms logically: $300 \times 0.0821 = 3 \times 8.21 = 24.63$. Then, multiplying by $0.5$ and dividing by $5$ is functionally equivalent to dividing directly by $10$, which shifts the decimal point one spot to the left: $2.463\ \mathrm{dm^3}$!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the mass ($W = 22\ \mathrm{g}$), temperature ($T = 300\ \mathrm{K}$), and pressure ($P = 5\ \mathrm{atm}$) of a carbon dioxide ($\mathrm{CO_2}$) gas sample. We need to determine the total volume ($V$) occupied by this gas under these state parameters, expressed in units of $\mathrm{dm^3}$.

Step 2: Key Formula or Approach:
Assuming ideal behavior, we apply the ideal gas equation: $$PV = nRT \implies V = \frac{nRT}{P}$$ Where the number of moles ($n$) is calculated from the given mass ($W$) and the molar mass ($M$) of the gas via $n = \frac{W}{M}$. Note that $1\ \mathrm{L} = 1\ \mathrm{dm^3}$.

Step 3: Detailed Explanation:
First, calculate the molecular weight of carbon dioxide ($\mathrm{CO_2}$): $$M = 12 + 2(16) = 44\ \mathrm{g\ mol^{-1}}$$ Now, determine the number of moles ($n$) present in the sample: $$n = \frac{22\ \mathrm{g}}{44\ \mathrm{g\ mol^{-1}}} = 0.5\ \mathrm{mol}$$ Substitute the parameters into the rearranged ideal gas law to solve for volume: $$V = \frac{0.5\ \mathrm{mol} \times 0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}} \times 300\ \mathrm{K}}{5\ \mathrm{atm}}$$ Simplify the expression: $$V = \frac{0.5 \times 300 \times 0.0821}{5} = \frac{150 \times 0.0821}{5}$$ $$V = 30 \times 0.0821 = 2.463\ \mathrm{L}$$ Since $1\ \mathrm{L} = 1\ \mathrm{dm^3}$, the volume equates to $2.46\ \mathrm{dm^3}$.

Step 4: Final Answer:
The volume occupied by the gas is $2.46\ \mathrm{dm^3}$, which matches option (C).