GASES: BOYLE’S LAW
A sample fluorine gas occupies a volume of 400 ml at 760 torr. Given that the temperature remains the same calculate the pressure required to reduce its volume by [tex]\frac{1}{2}[/tex].
GIVEN:
- [tex] \rm P_1 = 760\: torr[/tex]
- [tex]\rm V_1 = 400 ml[/tex]
- [tex]\rm V_2= \frac{1}{2}V_1[/tex]
SOLUTION:
Find the value of the final volume.
[tex]\rm V_2= \frac{1}{2}V_1 = \frac{1}{2} (400 \: ml) = 200 \: ml [/tex]
Rearrange the equation [tex]\rm P_1V_1 = P_2V_2[/tex] to get the equation that will solve for [tex]\rm P_2[/tex]
[tex]\rm P_2=\frac{P_1V_1}{V_2}[/tex]
Substitute all the given values to solve for the final temperature. The unit mL is cancelled out to give the final unit which is torr.
[tex]\rm P_2=\frac{P_1V_1}{V_2}[/tex]
[tex]\rm P_2=\frac{(760\:torr)(400\: mL)}{200\: mL}[/tex]
[tex]\rm P_2=\frac{304 \: 000 \: torr}{200}[/tex]
[tex] \boxed{\rm P_2=1 \: 520 \: torr}[/tex]
Convert the pressure in atmosphere, atm, using the equation [tex]\rm atm=\frac{torr}{760}[/tex]
[tex]\rm atm=\frac{torr}{760}[/tex]
[tex]\rm atm=\frac{1 \: 520}{760} = 2 [/tex]
- Using the same equation, we can say that the initial pressure is at one atmosphere (1 atm).
ANSWER:
Therefore, the pressure required to reduce its volume by [tex]\frac{1}{2}[/tex] is 2 atm.
