High School

A sample of gas initially at 4.00 atm was compressed from 8.00 L to 2.00 L at constant temperature. After the compression, what was the gas pressure in atm?

A) 2.00
B) 4.00
C) 8.00
D) 16.0
E) 1.00

Answer :

To solve this problem, we need to use Boyle's Law, which relates the pressure and volume of a gas at constant temperature. Boyle's Law is expressed by the equation:

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

Where:

  • [tex]P_1[/tex] is the initial pressure (4.00 atm).
  • [tex]V_1[/tex] is the initial volume (8.00 L).
  • [tex]P_2[/tex] is the final pressure (what we're solving for).
  • [tex]V_2[/tex] is the final volume (2.00 L).

Using Boyle's Law, we can solve for [tex]P_2[/tex]:

[tex]4.00 \, \text{atm} \times 8.00 \, \text{L} = P_2 \times 2.00 \, \text{L}[/tex]

Simplifying the equation:

[tex]32.00 \, \text{atm} \cdot \text{L} = P_2 \times 2.00 \, \text{L}[/tex]

Divide both sides by 2.00 L to solve for [tex]P_2[/tex]:

[tex]P_2 = \frac{32.00 \, \text{atm} \cdot \text{L}}{2.00 \, \text{L}} = 16.00 \, \text{atm}[/tex]

Therefore, the final pressure after the compression is 16.0 atm. This corresponds to option D in the multiple-choice answers.

So, the correct answer is: D) 16.0

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