Charles's law

Classical Mechanics Level 2

The temperature of an ideal gas held at constant pressure increases from 5 0 ∘ C 50^\circ \text{C} to 37 3 ∘ C . 373^\circ \text{C}. If the initial volume was 20 L , 20 \text{ L}, what is the final volume in L ? \text{L}?

40 10 149.2 20 2.68

July Thomas by July Thomas , 30, USA

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1 solution

July Thomas
Apr 30, 2016

Relevant wiki: Pressure

Charles's law is

V 1 T 1 = V 2 T 2 . \frac{V_1}{T_1}=\frac{V_2}{T_2}.

It is important to note that the temperatures are given in o C ^\text{o}\text{C} but Charles's law requires they be in K . \text{K}.

T 1 = 50 + 273 = 323 K T_1 = 50 + 273 = 323\text{ K}

T 2 = 373 + 273 = 646 K T_2 = 373 + 273 = 646\text{ K}

Combined with the provided V 1 = 20 L , V_1 = 20 \text{ L}, Charles's law is now

20 323 = V 2 646 \frac{20}{323} = \frac{V_2}{646}

V 2 = 40 L . V_2 = 40\text{ L}.

7 Helpful 3 Interesting 0 Brilliant 0 Confused

Hey , Can we also include Amanton's Law in the pressure wiki cause' the first 2 gas laws are already there and we can include that law and Avogadro's Law.

Abhiram Rao - 5 years, 1 month ago

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