The volume of a gas at 155.0 kPa changes from 22.0 L to 10.0 L. What
is the new pressure if the temperature remains constant?

V1P1=V2P2

155(22L)=P(10L0

P=341kPg

Explanation:

The following format is copied directly from your notes for how to solve these!!!

1) Analyze problem statement to find knowns and unknowns for each gas variable

V1 = 22.0 L P1 = 155.0 kPa T1 = T2 n = constant (same sample of gas)

V2= 10.0 L P2 = ? kPa T2 = T1 temperature is constant

2) Decide which of the gas laws to use and write its formula.

Only P and V are given, so Boyle's law is used P1V1 = P2V2

3) Change any temperature values to Kelvin (if T is needed) not needed

4) Plug in the knowns - INCLUDING UNITS!!

P1V1 = P2V2

(155.0 kPa) (22.0 L) = P2 (10.0 L)

P2 = (155.0 kPa)(22.0 L) = 341 kPa

(10.0 L)The following format is copied directly from your notes for how to solve these!!!

1) Analyze problem statement to find knowns and unknowns for each gas variable

V1 = 22.0 L P1 = 155.0 kPa T1 = T2 n = constant (same sample of gas)

V2= 10.0 L P2 = ? kPa T2 = T1 temperature is constant

2) Decide which of the gas laws to use and write its formula.

Only P and V are given, so Boyle's law is used P1V1 = P2V2

3) Change any temperature values to Kelvin (if T is needed) not needed

4) Plug in the knowns - INCLUDING UNITS!!

P1V1 = P2V2

(155.0 kPa) (22.0 L) = P2 (10.0 L)

P2 = (155.0 kPa)(22.0 L) = 341 kPa

(10.0 L)

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.