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A tank of bromine gas is initially at a pressure of 5.6 atm, a temperature of 67°C, and has a volume of 97 L. The tank is expanded to 100 L and has a new pressure of 8 atm.

What is the new temperature of the gas in Kelvin?

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nuhulawal20

The new temperature of the bromine gas as the tank is expanded to the given volume and pressure is 500.96 Kelvin.

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 in the question;

  • Initial volume V₁ = 97L
  • Initial pressure P₁ = 5.6atm
  • Initial temperature T₁ = 67°C = 340.15K
  • Final volume V₂ = 100L
  • Final pressure P₂ = 8.0atm
  • Final temperature T₂ = ?

To calculate the new temperature the gas, we subtsitute our given values into the expression above.

P₁V₁/T₁ = P₂V₂/T₂

P₁V₁T₂ = P₂V₂T₁

T₂ = P₂V₂T₁/P₁V₁

T₂ = ( 8.0atm × 100L × 340.15K ) / ( 5.6atm × 97L )

T₂ = 272120LatmK / 543.2Latm

T₂ = 500.96K

Therefore, the new temperature of the bromine gas as the tank is expanded to the given volume and pressure is 500.96 Kelvin.

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

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