Question

What pressure (in mmHg) is required to compress 15.6 liters of air at 1.20 atmospheres into a cylinder whose volume is 26.0 liters?

Answer 1

To compress the air from 15.6 L to 26.0 L, a pressure of approximately 547 mmHg is required.

Step-by-step explanation:

To solve this problem, we can use Boyle's law, which states that the pressure and volume of a gas are inversely proportional when temperature is constant.

Using the formula P1V1 = P2V2, we can plug in the given values:

P1 = 1.20 atm, V1 = 15.6 L, V2 = 26.0 L

Solving for P2, we can rearrange the equation to get:

P2 = (P1 * V1) / V2 = (1.20 atm * 15.6 L) / 26.0 L = 0.72 atm

To convert the pressure to mmHg, we can use the conversion factor 1 atm = 760 mmHg. Therefore, the pressure is:

P2 = 0.72 atm * 760 mmHg/atm ≈ 547 mmHg