Chemistry Reference
In-Depth Information
9.
[0.058 moles]—Work shown following:
Convert:
1 dm
3
1000 cm
3
550 cm
3
=
= 0.55 dm
3
Given: P = 98.4 kPa; V = 0.55 dm
3
; R = 8.31 dm
3
× kPa/mole × K;
T = 112 K
Find:n
PV
RT
=
(9
8
.4 kPa)(0.55 dm
3
)
(8.31 dm
3
× kPa/mol × K)(112 K)
= .0581485 = 0.058 moles
Formula:
n
=
10. [13 K]—Work shown following:
Convert:
1 kPa
7.50 mm of Hg
850 mm of Hg =
= 113 kPa
Given: P = 113 kPa; V = 4.5 dm
3
; n = 4.55 moles;
R = 8.31 dm
3
× kPa/mole × K
Find:
T
PV
nR
(113 kPa)(
4
.5 dm
3
)
(4.55 mol)(8.31 dm
3
× kPa/mol × K)
= 13.44864 K = 13 K
Formula:
T
=
=
Chapter 8 Examination
1. [b. Charles's Law]
2. [d. Graham's Law]
3. [c. Dalton's Law]
4. [a. Boyle's Law]
5. [e. Ideal Gas Law]
6. [d. Graham's Law]
7. [c. Dalton's Law]
8. [e. Ideal Gas Law]
9. [a. Boyle's Law]
10. [b. Charles's Law]
11. [B. high temperature and low pressure]—When the gas sample is at its
largest, the actual individual particle size will be of least significance. When
the particles are far apart and moving fast, the intermolecular forces will be
at their lowest.
12. [A. He]—Of the choices shown, helium will have the smallest particles and
the least intermolecular attraction between particles, making it the most
like an ideal gas.
