Question

boyles law - due 2-19-26
name
boyles law states that the volume of a gas varies inversely with its pressure if temperature
is held constant. (if one goes up, the other goes down.) we use the formula:
$p_1 \times v_1 = p_2 \times v_2$
solve the following problems (assuming constant temperature).

1. a sample of oxygen gas occupies a volume of 250. ml at 740. torr pressure. what

volume will it occupy at 800. torr pressure?

1. 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?

1. a 2.0 liter container of nitrogen had a pressure of 3.2 atm. what volume would be

necessary to decrease the pressure to 1.0 atm?

1. ammonia gas occupies a volume of 450. ml at a pressure of 720. mm hg. what

volume will it occupy at standard pressure?

1. a 175 ml sample of neon had its pressure changed from 75 kpa to 150 kpa.

what is its new volume?

1. a sample of hydrogen at 1.5 atm had its pressure decreased to 0.50 atm producing

a new volume of 750 ml. what was its original volume?

1. chlorine gas occupies a volume of 1.2 liters at 720 torr pressure. what volume will it

occupy at 1 atm pressure?

1. fluorine gas exerts a pressure of 900. torr. when the pressure is changed to 1.50 atm,

its volume is 250. ml. what was the original volume?

Explanation:

Step1: Rearrange Boyle's Law for $V_2$

$V_2 = \frac{P_1 V_1}{P_2}$

Step2: Substitute values for Q1

$V_2 = \frac{740. \text{ torr} \times 250. \text{ mL}}{800. \text{ torr}} = 231.25 \text{ mL}$

Step3: Rearrange Boyle's Law for $P_2$

$P_2 = \frac{P_1 V_1}{V_2}$

Step4: Substitute values for Q2

$P_2 = \frac{125 \text{ kPa} \times 3.50 \text{ L}}{2.00 \text{ L}} = 218.75 \text{ kPa}$

Step5: Substitute values for Q3

$V_2 = \frac{3.2 \text{ atm} \times 2.0 \text{ L}}{1.0 \text{ atm}} = 6.4 \text{ L}$

Step6: Use standard pressure (760 mm Hg) for Q4

$V_2 = \frac{720. \text{ mm Hg} \times 450. \text{ mL}}{760. \text{ mm Hg}} \approx 426.32 \text{ mL}$

Step7: Substitute values for Q5

$V_2 = \frac{75 \text{ kPa} \times 175 \text{ mL}}{150 \text{ kPa}} = 87.5 \text{ mL}$

Step8: Rearrange for original $V_1$ (Q6)

$V_1 = \frac{P_2 V_2}{P_1}$

Step9: Substitute values for Q6

$V_1 = \frac{0.50 \text{ atm} \times 750 \text{ mL}}{1.5 \text{ atm}} = 250 \text{ mL}$

Step10: Convert 1 atm to 760 torr (Q7)

$V_2 = \frac{720 \text{ torr} \times 1.2 \text{ L}}{760 \text{ torr}} \approx 1.13 \text{ L}$

Step11: Convert 1.50 atm to 1140 torr (Q8)

$V_1 = \frac{1140 \text{ torr} \times 250. \text{ mL}}{900. \text{ torr}} \approx 316.67 \text{ mL}$

Answer:

1. $231 \text{ mL}$ (rounded to 3 sig figs)
2. $219 \text{ kPa}$ (rounded to 3 sig figs)
3. $6.4 \text{ L}$
4. $426 \text{ mL}$ (rounded to 3 sig figs)
5. $88 \text{ mL}$ (rounded to 2 sig figs)
6. $250 \text{ mL}$
7. $1.1 \text{ L}$ (rounded to 2 sig figs)
8. $317 \text{ mL}$ (rounded to 3 sig figs)