Chemistry Reference
In-Depth Information
8.
[320 cm
3
]—Change 22.0
o
C to Kelvin and then solve, as shown below.
V
2
=
V
1
T
2
(350 cm
3
)(273 K)
(295 K)
=
= 323.898 cm
3
T
1
which rounds to 320 cm
3
9.
[546 K]—Standard temperature is 273 K. Solve as shown here.
V
2
T
1
V
1
(10.0 L)(273 K)
(5.00 K)
T
2
=
=
= 546 K
10. [0.805 L]—Remember to change the temperatures to K, by adding 273.
V
2
=
V
1
T
2
(0.855 L)(273 K)
(290. K)
=
= 0.804879 L
T
1
which rounds to 0.805 L.
Lesson 8-4 Review
1.
[1.7 atm]—P
He
= P
total
- P
H
= 2.5 atm - 0.8 atm = 1.7 atm
2.
[154.4 kPa]—P
total
= P
1
+ P
2
+ P
3
= 45.5 kPa + 23.6 kPa + 85.3 kPa
= 154.4 kPa
3.
[713.5 mm of Hg]—P
dry hydrogen
= P
total
- P
H
2
O
= 745.3 mm of Hg - 31.8 mm of
Hg (from the Vapor Pressure of Water table) = 713.5 mm of Hg
4.
[3.0 atm]—Gas is (2.0 moles/4.0 moles) = 50% oxygen.
6.0 atm × 0.50 = 3.0 atm
5.
[75.0 kPa]—P
total
= P
1
+ P
2
+ P
3
= 25.0 kPa + 25.0 kPa + 25.0 kPa
= 75.0 kPa
6.
[B. 1.3 atm]—P
Ar
= P
total
- P
Ne
= 1.8 atm - 0.5 atm = 1.3 atm
7.
[C. 83.3 kPa]—P
dry chlorine
= P
total
- P
H
2
O
= 85.6 kPa - 2.34 kPa = 83.26 kPa,
which rounds to 83.3 kPa
8.
[B. 0.200 atm]—Carbon dioxide makes up (1.00 mole/5.00 mole) = 20.0%
of gas. 1.0 atm × .200 = 0.200 atm
Lesson 8-5 Review
1.
[7.9 L]—Work shown following:
Given: V
1
= 7.6 L; T
1
= 296 K; P
1
= 1.12 atm; T
2
= 273 K; P
2
= 1.00 atm
Find: V
2
Formula:
V
1
P
1
T
2
P
2
T
1
(7.6 L)(1.12 atm)(273 K)
(1.00 atm)(296 K)
V
2
=
=
=7.85059 = 7.9 L
2.
[370. cm
3
]—Work shown following:
Given: V
1
= 250. cm
3
; T
1
= 273 K; P
1
= 101.3 kPa; T
2
= 359 K; P
2
= 89.9 kPa
Find: V
2
Formula:
V
1
P
1
T
2
P
2
T
1
(250. cm
3
)(101.3 kPa)(359 K)
(89.9 kPa)(273 K)
V
2
=
=
= 370.4431 cm
3
= 370. cm
3
