The result of increasing the Kelvin temperature by a factor of 2 at constant pressure and moles of the gas through the given diagrams I to IV has to be described. Concept Introduction: Charles’s Law tells that the relationship between temperature and volume of a gas. The law states that at constant pressure the volume of a fixed amount of gas is directly proportional to the temperature in Kelvin units. The mathematical expression for the relationship between volume and Kelvin temperature according to the Charles’s law can be shown as follows, V 1 T 1 = V 2 T 2 Where, V 1 = T h e v o l u m e o f a g a s a t a g i v e n p r e s s u r e . T 1 = T h e K e l v i n t e m p e r a t u r e o f t h e g a s . V 2 = T h e v o l u m e o f t h e g a s u n d e r a n e w s e t o f c o n d i t i o n s . T 2 = T h e K e l v i n t e m p e r a t u r e o f t h e g a s u n d e r a n e w s e t o f c o n d i t i o n s .

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Chapter 7, Problem 7.32EP

Interpretation Introduction

Interpretation:

The result of increasing the Kelvin temperature by a factor of 2 at constant pressure and moles of the gas through the given diagrams I to IV has to be described.

Concept Introduction:

Charles’s Law tells that the relationship between temperature and volume of a gas. The law states that at constant pressure the volume of a fixed amount of gas is directly proportional to the temperature in Kelvin units. The mathematical expression for the relationship between volume and Kelvin temperature according to the Charles’s law can be shown as follows,

V1T1=V2T2

Where,

V1 = The volume of a gas at a given pressure.T1 = The Kelvin temperature of the gas.V2 = The volume of the gas under a new set of conditions.T2 = The Kelvin temperature of the gas under a new set of conditions.

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