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Question
boyles law
boyles law states that the volume of a gas varies inversely with its pressure if temperature is held constant. (this means if one goes up, the other goes down.)
boyles law is: p1v1 = p2v2
solve the following problems (assume that temperature is constant).
- a sample of oxygen gas occupies a volume of 250. ml at 740. torr pressure. what volume will it occupy at 800. torr pressure?
- a sample of carbon dioxide occupies a volume of 3.50 liters at 125 kpa pressure. what pressure would the gas exert if the volume was decreased to 2.00 liters?
- a 2.0 - liter container of nitrogen had a pressure of 3.2 atm. what volume would be neces to decrease the pressure to 1.0 atm?
- ammonia gas occupies a volume of 450. ml at a pressure of 720. mmhg. what volur it occupy at standard pressure?
- a 175 ml sample of neon had its pressure changed from 75 kpa to 150 kpa. what new volume?
- chlorine gas occupies a volume of 1.2 liters at 720 torr pressure. wh will it occupy at 1 atm pressure?
Step1: Identify given values and formula
We use Boyle's Law $P_1V_1 = P_2V_2$.
Step2: Solve for unknown variable
For each problem, we re - arrange the formula $V_2=\frac{P_1V_1}{P_2}$ (when solving for volume) or $P_2=\frac{P_1V_1}{V_2}$ (when solving for pressure).
Problem 1
$P_1 = 740$ torr, $V_1=250$ mL, $P_2 = 800$ torr.
$V_2=\frac{P_1V_1}{P_2}=\frac{740\times250}{800}=231.25$ mL
Problem 2
$P_1 = 125$ kPa, $V_1 = 3.50$ L, $V_2=2.00$ L.
$P_2=\frac{P_1V_1}{V_2}=\frac{125\times3.50}{2.00}=218.75$ kPa
Problem 3
$P_1 = 3.2$ atm, $V_1 = 2.0$ L, $P_2 = 1.0$ atm.
$V_2=\frac{P_1V_1}{P_2}=\frac{3.2\times2.0}{1.0}=6.4$ L
Problem 4
Standard pressure $P_2 = 760$ mmHg, $P_1 = 720$ mmHg, $V_1 = 450$ mL.
$V_2=\frac{P_1V_1}{P_2}=\frac{720\times450}{760}\approx426.32$ mL
Problem 5
$P_1 = 75$ kPa, $V_1 = 175$ mL, $P_2 = 150$ kPa.
$V_2=\frac{P_1V_1}{P_2}=\frac{75\times175}{150}=87.5$ mL
Problem 6
$1$ atm = 760 torr, $P_1 = 720$ torr, $V_1 = 1.2$ L, $P_2 = 760$ torr.
$V_2=\frac{P_1V_1}{P_2}=\frac{720\times1.2}{760}\approx1.13$ L
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- $231.25$ mL
- $218.75$ kPa
- $6.4$ L
- $\approx426.32$ mL
- $87.5$ mL
- $\approx1.13$ L