To solve for the volume of the gas given the conditions specified, we can use the Ideal Gas Law, which is expressed as:
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm),
- [tex]\( V \)[/tex] is the volume in liters (L),
- [tex]\( n \)[/tex] is the number of moles of the gas,
- [tex]\( R \)[/tex] is the ideal gas constant ([tex]\( 0.0821 \frac{L \cdot atm}{mol \cdot K} \)[/tex]),
- [tex]\( T \)[/tex] is the temperature in Kelvin (K).
Given data:
- [tex]\( n = 0.223 \)[/tex] moles,
- [tex]\( T = 33.0 \)[/tex] degrees Celsius,
- [tex]\( P = 2.00 \)[/tex] atm.
First, we need to convert the temperature from degrees Celsius to Kelvin:
[tex]\[ T(K) = T(^{\circ}C) + 273.15 \][/tex]
[tex]\[ T(K) = 33.0 + 273.15 \][/tex]
[tex]\[ T(K) = 306.15 \, \text{K} \][/tex]
Next, we use the Ideal Gas Law to solve for [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
Substitute the given values into the equation:
[tex]\[ V = \frac{(0.223 \, \text{mol}) \times (0.0821 \, \frac{L \cdot atm}{mol \cdot K}) \times (306.15 \, \text{K})}{2.00 \, \text{atm}} \][/tex]
Now perform the calculations:
[tex]\[ V = \frac{0.223 \times 0.0821 \times 306.15}{2.00} \][/tex]
[tex]\[ V = \frac{5.60508665}{2.00} \][/tex]
[tex]\[ V \approx 2.8025430225 \, \text{L} \][/tex]
Rounding to two decimal places, we get:
[tex]\[ V \approx 2.80 \, \text{L} \][/tex]
Therefore, the volume of the gas is
[tex]\[
\boxed{2.80 \, \text{L}}
\][/tex]
Answer choice A) 2.80 L is the correct answer.
