Chapter 6 Thermodynamics: The First Law Worksheets Answers Page 2
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12Chapter 6
−2
g = 9.81 m⋅s
∆h = 15 m
−2
4
)(15 m) = 1.18 × 10
w = (80 kg)(9.81 m⋅s
J = 11.8 kJ
Example 6.2c
Describe the internal energy change and work performed when a spring is compressed or
expanded.
Answer
Work is done on the system by the surroundings to compress the spring. The compressed
spring has a greater capacity to do work than the uncompressed spring. In each case, the
internal energy of the system is increased by the amount of work done on it.
∆U > 0 and ∆U = U
− U
= w
final
initial
Therefore, w > 0 when work is done on the system.
Work is done on the surroundings by the system to expand the spring. The expanded spring
has a lesser capacity to do work than the unexpanded spring. The internal energy of the
system is decreased by the amount of work done by it.
∆U < 0 and ∆U = U
− U
= w
final
initial
Therefore, w < 0 when work is done by the system.
E xample 6.2d
Describe the internal energy change and work performed when a battery is recharged.
Answer
Electrical work is done on the system by the surroundings in recharging a battery. The
charged battery has a greater capacity to do work than the discharged battery. The internal
energy of the system is increased by the amount of work done on it. As in the case of
mechanical work, w > 0 when electrical work is done on the system.
6.3 Expansion Work
Example 6.3a
The pressure exerted on an ideal gas at 2.00 atm and 300 K is reduced suddenly to 1.00 atm
while heat is transferred to maintain the initial temperature of 300 K. Calculate q, w, and
∆U in joules for this process.
For an ideal gas undergoing an isothermal process, ∆U = 0. The work done at constant
Solution
external pressure is w = − P
∆V, where ∆V = V
− V
.
ex
final
initial
−1
−1
⋅mol
V
= (nRT )/P
= (1 mol)(0.082 06 L⋅atm⋅K
)(300 K)/(2.00 atm) = 12.3 L
initial
initial
−1
−1
⋅mol
V
= (nRT )/P
= (1 mol)(0.082 06 L⋅atm⋅K
)(300 K)/(1.00 atm) = 24.6 L
final
final
∆V = V
− V
= 24.6 L − 12.3 L = 12.3 L (system expands)
final
initial
w = − P
∆V = −(1.00 atm)(12.3 L) = −12.3 L⋅atm
ex
−1
−1
⋅atm
) = −1.25 × 10
3
= (−12.3 L⋅atm)(101.325 J⋅L
J
Work is done by the system on the surroundings.
q = ∆U − w = 0 − (−1.25 × 10
3
J) = +1.25 × 10
3
J
Heat is absorbed by the system from the surroundings.
Example 6.3b
Suppose the pressure change in Example 6.3a is carried out reversibly as well as
isothermally. Calculate q, w, and ∆U in joules for this process and compare the answers
to those in the irreversible expansion in one step.
The internal energy of an ideal gas depends only on temperature, and ∆U = 0.
Answer
Check the two formulas for isothermal, reversible work.
V
24.6 L
−
−
1
1
= −
final
= −
⋅
⋅
w
nRT
ln
(1.00 mol)( 8.314 47 J K
mol )(300 K) ln
V
12.3 L
initial
2
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