A container with a volume of 2.0 L is filled with a gas at a pressure of 1.5 atm. By decreasing the volume of the container to 1.0 L, what is the resulting pressure?

Use Boyle's law: [tex]\( P_1 V_1 = P_2 V_2 \)[/tex]

Provide your answer with the correct number of significant figures in atm.



Answer :

To find the resulting pressure when the volume of the container is decreased, we can use Boyle's Law, which states that for a given amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume. Mathematically, Boyle's Law is expressed as:

[tex]\[ P_1 \cdot V_1 = P_2 \cdot V_2 \][/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, and
- [tex]\( V_2 \)[/tex] is the final volume.

Given the values:
- [tex]\( V_1 = 2.0 \, L \)[/tex]
- [tex]\( P_1 = 1.5 \, atm \)[/tex]
- [tex]\( V_2 = 1.0 \, L \)[/tex]

We need to find the final pressure [tex]\( P_2 \)[/tex]. Let's rearrange the formula to solve for [tex]\( P_2 \)[/tex]:

[tex]\[ P_2 = \frac{P_1 \cdot V_1}{V_2} \][/tex]

Now, substituting the known values into the equation:

[tex]\[ P_2 = \frac{1.5 \, \text{atm} \cdot 2.0 \, \text{L}}{1.0 \, \text{L}} \][/tex]

Perform the multiplication and division:

[tex]\[ P_2 = \frac{3.0 \, \text{atm} \cdot \text{L}}{1.0 \, \text{L}} \][/tex]

[tex]\[ P_2 = 3.0 \, \text{atm} \][/tex]

Therefore, the resulting pressure when the volume of the container is decreased to [tex]\( 1.0 \, L \)[/tex] is [tex]\( 3.0 \, \text{atm} \)[/tex], using the correct number of significant figures.