Gas Law Problems
2. If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its
volume at a pressure of 2.50 atm? (Use Boyle’s Law)
P1V1 = P2V2
P1=1.00 atm
V1= 3.60 L
P2= 2.5 atm
V2=?
(1.00 atm) (3.60 L) = (2.5 atm) ( V2)
3.60 atm L = 2.5 atm V2
2.5 atm
2.5 atm
V2 = 1.44 L
3. A balloon is inflated to 5L at 20°C. What will be the new volume of the balloon if
it is put in a freezer at 0°C? . (Use Charles’ Law)
V1 = V2
T1 = T2
V1= 5 L
T1= 293 K
V2= ?
T2= 273 K
273 K 5L
= V2
(273 K)
293 K
273 K
1365 L = V2
293
V2 = 4.66 L
4. A gas occupies 1.56 L at 1.00 atm. What will be the volume of this gas if the
pressure becomes 3.00 atm? (Use Boyle’s Law)
P1V1 = P2V2
P1=1.00 atm
V1= 1.56 L
(1.00 atm) (1.56 L) = (3 atm) ( V2)
1.56 atm L = 3 atm V2
3 atm
3 atm
V2 = 0.52 L
P2= 3 atm
V2=?
Gas Law Problems
5. A cylinder of chlorine gas is stored in a concrete-lined room for safety. The
cylinder is designed to withstand 50.0atm of pressure. The pressure gage reads
35.0 atm at 23.2C. An accidental fire in the room next door causes the temperature
in the storeroom to increase to 87.5C. What will the pressure gage read at this
temperature? (Use Gay-Lussac’s Law)
P1 / T1 = P2 / T2
P1 = 35 atm
T1 = 23.2 C
P2 = ?
T2 = 87.5C
35 atm / 23.2*C = P2 / 87.5*C
(87.5*C) 35 atm / 23.2*C = P2 / 87.5*C (87.5*C)
3062.5 atm / 23.2 = P2
P2 = 132.00 atm
6. A gas occupies 11.2 liters at 0.860 atm. What is the pressure if the volume
becomes 15.0 L? (Use Boyle’s Law)
P1V1 = P2V2
P1 = 0.86 atm
V1 = 11.2 L
P2 = ?
V2= 15 L
(0.86 atm) (11.2 L) = (P2) ( 15 L)
15L
15 L
P2 = 0.642 atm
7. If you took a balloon outside that was originally at 20°C at 2 L in volume, and it
heated up to 29°C, then what would its new volume be? (Use Charles’ Law)
V1 = V2
T1 = T2
V1= 2 L
T1= 293 K
302 K 2L
= V2
(302 K)
293 K
302 K
604 L = V2
293
V2 = 2.06 L
V2= ?
T2= 302 K
Gas Law Problems
8. A 40.0 L tank of ammonia has a pressure of 1000 mmHg. Calculate the volume of
the ammonia if its pressure is changed to 500 torr while its temperature remains
constant.
P1V1 = P2V2
P1 = 1000 torr
V1 = 40 L
P2 = 500 torr
V2= ?
(1000 torr) (40 L) = (V2) ( 500 torr)
500 torr
500 torr
40000 L = V2
500
V2 = 80 L
9. Chlorine gas occupies a volume of 1.2 liters at 720 torr pressure. What volume
will it occupy at 1 atm pressure? (Use Boyle’s Law)
P1V1 = P2V2
P1 = 1 atm
V1 = 1.2 L
P2 = 0.95 atm
V2= ?
(1 atm) (40 L) = (V2) ( 0.95 atm)
0.95 atm
0.95 atm
40 L = V2
0.95
V2 = 42.11 L
1 atm = 101.3 kPa = 101,325 Pa = 760 mm Hg = 760 torr
Use these conversion factors (dimensional analysis) to solve 8 and 9.
Gas Law Problems