Concept explainers
(a)
Interpretation:
The following table is to be completed by assuming the gas at a constant volume.
3.25atm | 298K | ? | 398K |
Concept Introduction:
According to Gay Lussac's law, the pressure of the gas is directly proportional to the Kelvin temperature of the gas at constant volume. This relation is represented as
Or,

Answer to Problem 39P
5.0L | 310K | 4.34atm | 250K |
Explanation of Solution
According to Gay Lussac's law, the pressure of the gas is directly proportional to the Kelvin temperature of the gas at constant volume. This relation is represented as
For (a) −
Given that −
Initial pressure, P1 = 3.25atm
Initial temperature, T1 = 298K
Final temperature, T2 = 398K
Put the above values in equation (1)
The final pressure of gas that is P2 = 4.34atm
Thus,
5.0L | 310K | 4.34atm | 250K |
(b)
Interpretation:
The following table is to be completed by assuming the gas at constant volume.
550mmHg | 273K | ? |
Concept Introduction:
According to Gay Lussac's law, the pressure of the gas is directly proportional to the Kelvin temperature of the gas at constant volume. This relation is represented as
Or,

Answer to Problem 39P
150mL | 45K | 349mmHg |
Explanation of Solution
According to Gay Lussac's law, the pressure of the gas is directly proportional to the Kelvin temperature of the gas at constant volume. This relation is represented as
Given that −
Initial pressure, P1 = 550mmHg
Initial temperature, T1 = 273K
Final temperature, T2 =
Or,
Put the above values in equation (1)
The final pressure of gas that is P2 = 349mmHg
150mL | 45K | 349mmHg |
(c)
Interpretation:
The following table is to be completed by assuming the gas at a constant volume.
0.25atm | 955mmHg | ? |
Concept Introduction:
The gas at constant volume. This relation is represented as
Or,

Answer to Problem 39P
60.0L | 180L | 1300K |
Explanation of Solution
According to Gay Lussac's law, the pressure of a gas is directly proportional to the Kelvin temperature of the gas at constant volume. This relation is represented as
Given that −
Initial volume, P1 = 0.50atm
Or,
1 atm = 760mm Hg
Final temperature, T1 =
Final volume, P2 = 955mmHg
Put the above values in equation (2)
The final temperature of the gas that is T2 = 1300K
60.0L | 180L | 1300K |
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Chapter 7 Solutions
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