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: Collect the data.
Mass $W = 22\ g$ of $CO_2$, temperature $T = 300\ K$, pressure $P = 5\ atm$, and $R = 0.0821\ L\ atm\ K^{-1}\ mol^{-1}$.
Step 2: Find moles of gas.
Molar mass of $CO_2 = 12 + 2(16) = 44\ g\ mol^{-1}$, so $n = \dfrac{22}{44} = 0.5\ mol$.
Step 3: Choose the working equation.
The ideal gas law gives $V = \dfrac{nRT}{P}$.
Step 4: Substitute the numbers.
\[ V = \frac{0.5 \times 0.0821 \times 300}{5}. \]
Step 5: Simplify step by step.
The numerator is $0.5 \times 300 = 150$, then $150 \times 0.0821 = 12.315$, and dividing by $5$ gives $V = 2.463\ L$.
Step 6: Convert and conclude.
Since $1\ L = 1\ dm^3$, the volume is about $2.46\ dm^3$, option (C).
\[ \boxed{2.46\ dm^3 \text{ (option C)}} \]