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

Answer 1

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

= 130^oC

T2 = 130^oC

Therefore, To 130^oC 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.