33) (a) A piston at 5.9 atm contains a gas that occupies a volume of 3.5 L. What
ID: 905146 • Letter: 3
Question
33)
(a) A piston at 5.9 atm contains a gas that occupies a volume of 3.5 L. What pressure would have to be placed on the piston to force the volume to adjust to 0.32 L?
_____________ atm
(b) A piston at -44.0°C contains a gas that occupies a volume of 2.5 L. To what temperature would the gas have to be heated to increase the volume to 4.4 L at constant pressure?
__________ °C
(c) A piston at 795 torr contains a gas that occupies a volume of 4.7 L. What pressure would have to be placed on the piston to force the volume to adjust to 0.78 L?
___________atm
Explanation / Answer
a) Initial pressure P1 = 5.9 atm
Initial volume V1 = 3.5 L
Final pressure P2 = ?
Final volume V2 = 0.32 L
From Boyle's law,
P1V1 = P2V2
P2 = P1V1/ V2
= (5.9 atm x 3.5 L) / 0.32 L
= 64.53 atm
P2 = 64.53 atm
Therefore, 64.53 atm pressure would have to be placed on the piston to force the volume to adjust to 0.32 L.
b)
Initial temperature T1 = -44 o C = -44 + 273 K = 229 K
Initial volume V1 = 2.5 L
Final temperature T2 = ?
Final volume V2 = 4.4 L
From Charles' law,
V1/T1 = V2/T2
T2 = (V2/ V1) T1
= (4.4 L/2.5 L) (229 K)
= 403.04 K
= 403.04 - 273 o C
= 130oC
T2 = 130oC
Therefore, To 130oC temperature would the gas have to be heated to increase the volume to 4.4 L at constant pressure.
c)
Initial pressure P1 = 795 torr = 795/760 atm ( 1 atm = 760 torr)
Initial volume V1 = 4.7 L
Final pressure P2 = ?
Final volume V2 = 0.78 L
From Boyle's law,
P1V1 = P2V2
P2 = P1V1/ V2
= [(795/760 atm) x 4.7 L] / 0.78 L
= 6.303 atm
P2 = 6.303 atm
Therefore, 6.303 atm pressure would have to be placed on the piston to force the volume to adjust to 0.78 L.