A 3.0 L container holds a sample of hydrogen gas at 150 kPa. The pressure increases to 2 atm, and the temperature remains constant.

What will the volume be?

A. 0.22 L
B. 0.44 L
C. 2.25 L
D. 4.50 L



Answer :

To determine the final volume of the gas in the container, we'll use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when the temperature remains constant. Mathematically, Boyle's Law is expressed as:

[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]

where:
- [tex]\( P_1 \)[/tex] is the initial pressure,
- [tex]\( V_1 \)[/tex] is the initial volume,
- [tex]\( P_2 \)[/tex] is the final pressure,
- [tex]\( V_2 \)[/tex] is the final volume.

We are given the following:
- Initial volume [tex]\( V_1 = 3.0 \text{ L} \)[/tex]
- Initial pressure [tex]\( P_1 = 150 \text{ kPa} \)[/tex]
- Final pressure [tex]\( P_2 = 2 \text{ atm} \)[/tex]

However, to use the formula properly, the pressures must be in the same units. We convert the initial pressure from kilopascals (kPa) to atmospheres (atm). The conversion factor is:

[tex]\[ 1 \text{ atm} = 101.325 \text{ kPa} \][/tex]

Thus,

[tex]\[ P_1 = \frac{150 \text{ kPa}}{101.325 \text{ kPa/atm}} \approx 1.480 \text{ atm} \][/tex]

Now we can use Boyle's Law:

[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]

Substitute the known values:

[tex]\[ 1.480 \text{ atm} \times 3.0 \text{ L} = 2 \text{ atm} \times V_2 \][/tex]

Solving for [tex]\( V_2 \)[/tex]:

[tex]\[ V_2 = \frac{1.480 \text{ atm} \times 3.0 \text{ L}}{2 \text{ atm}} \][/tex]

[tex]\[ V_2 \approx \frac{4.440 \text{ atm} \cdot \text{L}}{2 \text{ atm}} \][/tex]

[tex]\[ V_2 \approx 2.220 \text{ L} \][/tex]

Therefore, the final volume of the gas when the pressure increases to [tex]\( 2 \text{ atm} \)[/tex] is approximately [tex]\( 2.22 \text{ L} \)[/tex]. Out of the given choices, the closest one is:

[tex]\[ 2.25 \text{L} \][/tex]