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The volume of a gas is [tex]$0.450 \, L[tex]$[/tex] when its pressure is [tex]$[/tex]1.00 \, atm[tex]$[/tex]. If the temperature of the gas does not change, what is the pressure when its volume is changed to [tex]$[/tex]2.00 \, L$[/tex]?

Use [tex]P_1 V_1 = P_2 V_2[/tex].

A. [tex]0.225 \, atm[/tex]
B. [tex]0.444 \, atm[/tex]
C. [tex]2.25 \, atm[/tex]
D. [tex]4.44 \, atm[/tex]



Answer :

GinnyAnswer
Sure, let's solve this problem step-by-step using Boyle's Law, which states that for a given mass of gas at constant temperature, the pressure and volume are inversely proportional. Mathematically, it can be expressed as:

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

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

Given:
- \( V_1 = 0.450 \) L
- \( P_1 = 1.00 \) atm
- \( V_2 = 2.00 \) L

We need to find the final pressure \( P_2 \).

First, rewrite the equation to solve for \( P_2 \):

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

Now, plug in the given values:

[tex]\[ P_2 = \frac{1.00 \text{ atm} \times 0.450 \text{ L}}{2.00 \text{ L}} \][/tex]

Perform the multiplication and division:

[tex]\[ P_2 = \frac{0.450 \text{ atm} \cdot \text{L}}{2.00 \text{ L}} \][/tex]

[tex]\[ P_2 = 0.225 \text{ atm} \][/tex]

So, the pressure when the volume is changed to \( 2.00 \) L is \( 0.225 \) atm.

Therefore, the correct answer is:

[tex]\[ 0.225 \text{ atm} \][/tex]

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