BOYLE’S LAW
Problem Solving
Use this equation:
P1V1 = P2V2
Problem #1:
2.00 L of a gas is at 740.0
mmHg pressure. What is
its volume at standard
pressure?
Problem #2:
10.0 L of a gas is at 2.16
atm. What pressure is
obtained when the
volume is 20.0 L?
Problem #3:
A gas occupies 24. 6 liters
at a pressure of 80.0 mmHg.
What is the volume when
the pressure is increased to
120 mmHg?
Problem #4:
If a gas at 50.0 ̊C occupies
5.50 liters at a pressure of
2.00 atm, what will be its
volume at a pressure of
4.50 atm?
Problem #5:
A gas occupies 11.2 liters
at 0.860 atm. What will
be the pressure if the
volume becomes 15.0 L?
ANSWERS
Problem #1: 3.00 L of a gas is at 30.0
mmHg pressure. What is its volume at
standard pressure of 760.0 mmHg?
1) Use this equation:
P1V1 = P2V2
2)Insert values:
(30.0 mmHg) (3.00 L) = (760.0 mmHg) (x)
3) Multiply the left side and divide (by 760.0 mmHg) to
solve for x.
x = 0.118 L (to three significant figures)
Problem #2: 10.0 L of a gas is at 2.16
atm. What pressure is obtained when
the volume is 20.0 L?
Solution:
Use the same technique as in Example #1:
(2.16 atm) (10.0 L) = (x) (20.0 L)
x = 1.08 atm (to three sig figs)
Problem #3: A gas occupies 24. 6 liters at
a pressure of 80.0 mmHg. What is the
volume when the pressure is increased to
120 mmHg?
Solution:
(80.0 mmHg) (24.6 liters) = (120 mmHg) (x)
X= 16.4 L
(Note three significant figures)
Problem #4: If a gas at 50.0 ̊C
occupies 5.50 liters at a pressure of
2.00 atm, what will be its volume at a
pressure of 4.50 atm?
Solution:
(2.00 atm) (5.50 liters) = (4.50 atm) (x)
X= 2.44 L
Problem #5: A gas occupies 11.2 liters
at 0.860 atm. What will be the
pressure if the volume becomes 15.0
L?
Solution:
(11.2 liters) (0.860 atm) = (x) (15.0 L)
X= 0.642 atm