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A gas at 1.10 atm and 30.0°C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0°C and the pressure increased to 2.40 atm, what is the new volume?

A gas at 110 atm and 300C fills a flexible container with an initial volume of 200 L If the temperature is raised to 800C and the pressure increased to 240 atm class=

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uanseojefaith

Answer:

P1=1.10atm

V1=2.00L

T1=30.0+273=303

P2=2.40atm

V2=?

T2=80.0+273=353

Explanation:

V2=1.10*2.00*353=776.6

2.40*303=727.2

776.6/727.2=1.0679L

nuhulawal20

The new volume occupied by the of gas as its temeprature and pressure is raised is 1.07L.

Combined gas law

Combined gas law put together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.

It is expressed as;

P₁V₁/T₁ = P₂V₂/T₂

Given the data question;

  • Initila pressure P₁ = 1.10atm
  • Final pressure p₂ = 2.40atm
  • Inital temperature T₁ = 30.0°C = 303.15K
  • Final temperature T₂ = 80.0°C = 353.15K
  • Initial Volume V₁ = 2.00L
  • Final Volume V₂ = ?

To calculate the new volume that the gas occupy, we subtsitute our given values into the expression above.

P₁V₁/T₁ = P₂V₂/T₂

P₁V₁T₂ = P₂V₂T₁

V₂ = P₁V₁T₂  / P₂T₁

V₂ = ( 1.10atm × 2.00L × 353.15K )  / ( 2.40atm × 303.15K )

V₂ = 776.93LatmK / 727.56atmK

V₂ = 1.07L

Therefore, The new volume occupied by the of gas as its temeprature and pressure is raised is 1.07L.

Learn more about the combined gas law here: brainly.com/question/25944795

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