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y = mx + b P = (constant)(1/V) + 0 PV = constant (answer to both 1 and 2)

y = mx + b P = (constant)(1/V) + 0 PV = constant (answer to both 1 and 2). Conclusions and Questions. 1. If P doubles, V is cut in half. If P is tripled, V becomes 1/3 of its original value.

candice-chaney
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y = mx + b P = (constant)(1/V) + 0 PV = constant (answer to both 1 and 2)

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  1. y = mx + b P = (constant)(1/V) + 0 PV = constant (answer to both 1 and 2)

  2. Conclusions and Questions 1. If P doubles, V is cut in half. If P is tripled, V becomes 1/3 of its original value. 2. “As the pressure on a gas increases, the volume of the gas decreases proportionally, provided that the temperature and amount of gas remain constant”. I.e. P1V1 = P2V2.

  3. 3. As volume decreases, there are more collisions with the side of the container per unit of time, thus the pressure increases. 4. Boyle’s states that pressure is inversely related to volume, if the temperature and # of particles (in moles, n) are constant

  4. By comparing values in the PV column and via the graph we see that PV = constant • Boyle’s law: P1V1 = P2V2

  5. Sample Questions 1. At 103 kPa, a gas has a volume of 5.2 L. What is its volume if the pressure is increased to 400 kPa? 2. A gas at 700 mm Hg has a volume of 200 mL. ? atm of pressure will make the volume 950 mL.

  6. Answers: 1. P1=103kPa, V1=5.2L, P2=400kPa, V2=? Using Boyle’s law: P1V1 = P2V2, 103 kPa x 5.2 L = 400 kPa x V2 V2= (103 kPa x 5.2 L) / 400 kPa = 1.34 L

  7. 2. P1= 700 mm Hg x 1 atm / 760 mm Hg = 0.921 atm, V1= 200 ml, P2= ?, V2= 950 ml Using Boyle’s law: P1V1 = P2V2 0.921 atm x 200 ml = P2 x 950 ml P2= (0.921 atm x 200 ml)/950ml=0.194 atm

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