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A gas is heated from 213.0 K to 398.0 K and the volume is increased from 13.0 liters to 35.0 liters by moving a large piston within a cylinder. If the original pressure was 3.15 atm, what would the final pressure be?

Respuesta :

RevyBreeze

Answer:

P₂ = 0.13880 atm

Explanation:

Given info are;

Initial volume = 13.0 L

Initial pressure = 3.15 atm

Initial temperature = 213.0 K

Final temperature = 398.0 K

Final volume = 35.0 L

Final pressure = ?

Note;

Base on general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Formula = P₁V₁/T₁ = P₂V₂/T₂  

  • P₁ = Initial pressure
  • V₁ = Initial volume
  • T₁ = Initial temperature
  • P₂ = Final pressure
  • V₂ = Final volume
  • T₂ = Final temperature

Solution:

P₂ = P₁V₁ T₂/ T₁ V₂

P₂ = 3.15 atm x 13.0 L x 398.0k/213.0 k x 35.0 L  

P₂ = 1034.8 atm .L.K/ 7455 K.L

P₂ = 0.13880617035

P₂ = 0.13880 atm  

When we multiply or divide the values the number of significant figures must be equal to the less number of significant figures in given value.

Thus we will include five significant figures in answer because 1034.8 have five significant figure.

[RevyBreeze]

Answer:

I suggest that you try this yourself.

Explanation:

Your online chemistry teacher knows that you are using this to answer the question.  

-From said teacher

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