Physics Lesson 13.9.3 - Relationship between Volume and Temperature at Constant Pressure. The Charles's Law

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Welcome to our Physics lesson on Relationship between Volume and Temperature at Constant Pressure. The Charles's Law, this is the third lesson of our suite of physics lessons covering the topic of Gas Laws, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Relationship between Volume and Temperature at Constant Pressure. The Charles's Law

If we supply some heat energy to the ideal gas inside the cylinder of the previous paragraphs by allowing the piston to slide freely up or down (here, up), we obtain a process at constant pressure as it is self-regulated automatically in order to balance the atmospheric pressure which acts from above the piston.

Physics Tutorials: This image provides visual information for the physics tutorial Gas Laws

Such a process - which occurs at constant pressure - is known as isobaric process. We write P₁ = P₂ = P. If we apply again the ideal gas law for the two states 1 and 2, we obtain

P × V₁ = n × R × T₁

for the initial state and

P × V₂ = n × R × T₂

for the final state of the gas, where n is the number of moles and R is the ideal gas constant (both remain unchanged during the entire process).

Dividing both equations side by side, we have

P × V₁/P × V₂ = n × R × T₁/n × R × T₂

V₁/V₂ = T₁/T₂

The above equation is known as the Charles's Law.

The corresponding P - V graph for an isovolumetric process is shown below.

The area under the graph represents the work done by the gas to lift the piston, as explained in the previous tutorial.

Example 4

An ideal gas contracts at constant pressure from 8 L to 3 L when some heat energy is taken off from it. As a result, the gas temperature becomes -3°C at the end of process. What w the initial temperature of gas in Celsius degree?

Solution 4

We have the following clues:

P₁ = P₂ = P
V₁ = 8 L = 0.008 m³
V₂ = 3 L = 0.003 m³
T₂ = - 23°C = - 3 + 273 = 270 K
T₁ = ?

Since this is an isobaric process, we apply the Charles's Law to find the initial temperature of gas. We have

V₁/V₂ = T₁/T₂
0.008 m³/0.003 m³ = T₁/270 K
T₁ = 0.008 m³ × 270 K/0.003 m³
= 720 K

When converted into Celsius scale, this value becomes 720 - 273 = 447°C.

You have reached the end of Physics lesson 13.9.3 Relationship between Volume and Temperature at Constant Pressure. The Charles's Law. There are 4 lessons in this physics tutorial covering Gas Laws, you can access all the lessons from this tutorial below.

More Gas Laws Lessons and Learning Resources

Thermodynamics Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
13.9	Gas Laws
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
13.9.1	Relationship between Pressure and Volume at Constant Temperature. The Boyle's Law
13.9.2	Relationship between Pressure and Temperature at Constant Volume. The Gay-Lussacs's Law
13.9.3	Relationship between Volume and Temperature at Constant Pressure. The Charles's Law
13.9.4	Relationship between the Volume of Ideal Gas and the Number of Molecules. The Avogadro's Law

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