Anonymous
Anonymous asked in Science & MathematicsChemistry · 9 years ago

boyle's law question?

use boyles law for the following calculation. a sample of gas which occupies a volume of 22.4 ml at a pressure of 1.21 atm is subjected to a decrease in pressure of 82 mmHg. calculate the new volume show work

Relevance
• 9 years ago

P1V1 = P2V2

22.4mL = .0224 L

82 mmHg = .11 atm, since 760mmHg = 1atm

(1.21atm)(.0224L) = (.11 atm)V

V = 0.2464 = .25 L

• 9 years ago

Ideal gas law: PV=nRT where P is pressure, V is volumes, n is amount of substance of gas, R is the ideal gas constant, and T is temperature. Measured in SI units, n is measured in moles, T in kelvin, and R has a value of 8.314 J/(K·mol).

Since you are dealing with only volume and pressure and the amount of the substance and temperature remain the same, P₁V₁=nRT and P₂V₂=nRT, thus P₁V₁=P₂V₂.

P₁V₁=P₂V₂

Plug in your givens. The gas initially has a volume of 22.4 mL and a pressure of 1.21 atm. After the decrease, it has a pressure of 82 mmHg. So, V₂ is the unknown variable.

(1.21 atm)(22.4 mL)=(82 mmHg)V₂

Another problem arises. The units of pressure on the left don't match the units of pressure on the right, so you have to convert either one to match the other. I suggest converting mmHg to atm. 82 mmHg * (1 atm/(759.99 mmHg)) = .11 atm

Plug that in for 82 mmHg.

(1.21 atm)(22.4 mL)=(.11 atm)V₂

Solve for V₂. Divide by .11 atm.

(1.21 atm)(22.4 mL)/(.11 atm)=V₂

Simplify.

246.4 mL=V₂

V₂=246.4 mL

The reason I use the ideal gas law instead of boyle's law is because you can easily remove whichever variables from the equation that don't change. For example, P and V may remain constant but n and T might change. You don't have to remember multiple equations, just this one.