Charles Law Solved Examples
Example 1: At 19°C, a sample of gas takes up 2.55 L. What is the new volume of the gas if the pressure stays the same and the temperature is increased to 30°C?
Solution:
T1 = 273 + 19 = 292 K
T2 = 273 + 30 = 303 K
Given,
- Initial Temperature = T1 = 292 K
- Initial Volume = V1 = 2.55 l
- Final Temperature = 303 K
- Final Volume = V2 = ?
Now, according to Charles’s Law
V1/T1 = V2/T2
2.55 / 292 = V2 /303
V2 = 2.55 × 303 / 292
V2 = 772.65 / 292
V2 = 2.64 L
Final Volume = V2 = 2.64 l
Example 2: At 123 degrees Celsius, helium gas has a volume of 1690 ml. Determine the temperature at which the capacity expands to 472 ml. Count on pressure remaining steady.
Solution:
T1 = 273 + 123 = 396 K
Given,
- Initial Temperature = T1 = 396 K
- Initial Volume = V1 = 1690 ml
- Final Temperature = ?
- Final Volume = V2 = 472 ml
Now, according to Charles’s Law
V1/T1 = V2/T2
V1/T1V2 = 1/T2
T2 = T1V2 / V1
T2 = 396 × 1690 / 472
T2 = 669240 / 472
T2 = 1417.9 K
T2 = 1417.9 – 273 = 1144.9°C
Final Temperature = T2 = 1144.9°C
Example 3: What will the initial gas volume be at 400 K if the final volume is 13 L at 270 K?
Solution:
Given,
- Initial Temperature = T1 = 400 K
- Initial Volume = V1 = ?
- Final Temperature = 270 K
- Final Volume = V2 = 13 l
Now, according to Charles’s Law
V1/T1 = V2/T2
V1 = V2T1 / T2
V1 = 13 × 400 / 270
V1 = 5200 / 270
V1 = 19.26 L
Initial Volume = V1 = 19.26 L
Example 4: What will happen to the gas’s volume if 55 mL of it is cooled from 67°C to 30°C?
Solution:
T1 = 273 + 67 = 340 K
T2 = 273 + 30 = 303 K
Given,
- Initial Temperature = T1 = 340 k
- Initial Volume = V1 = 55 ml
- Final Temperature = 303 K
- Final Volume = V2 = ?
Now, according to Charles’s Law
V1/T1 = V2/T2
V2 = V1T2 / T1
V2 = 55 × 303 / 340
V2 = 16665 /340
V2 = 49.015 ml
Final Volume = V2 = 49.015 ml
Example 5: A gas has a volume of 1.5 L at normal temperature. At a temperature of 333 K, determine the volume.
Solution:
Given,
- Initial Temperature = T1 = 273 k
- Initial Volume = V1 = 1.5 l
- Final Temperature = T2 = 333k
- Final Volume = V2 = ?
Now, according to Charles’s Law
V1/T1 = V2/T2
V2 = V1T2 / T1
V2 = 1.5 × 333 / 273
V2 = 499.5 / 273
V2 = 1.8 L
Final Volume = V2 = 1.8 L
Example 7: At standard temperature the volume is 30 L, find the final temperature has a volume of 50 L.
Solution:
Given,
- Initial Temperature = T1 = 273 k
- Initial Volume = V1 = 30 l
- Final Temperature = ?
- Final Volume = V2 = 50 l
Now, according to Charles’s Law
V1/T1 = V2/T2
T2 = T1V2 / V1
T2 = 273 × 30 / 50
T2 = 8190 / 50
T2 = 163.8 K
Final Temperature = T2 = 163.8 K
Related Article,
Charles Law
Charles Law is one of the fundamental laws used for the study of gases. Charles Law states that the volume of an ideal gas is directly proportional to the temperature (at absolute scale) at constant pressure. Famous French physicist Jacques Charles formulated this law in the year 1780.
Let’s learn about Charles Law’s derivation and others in detail in this article.
Table of Content
- What is Charles Law?
- Jacques Charles
- Charles’s Law Formula
- Derivation of Charles Law Formula
- Experiment Verification of Charles law
- Limitations of Charles law
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