Question

A gas has a pressure of 120 kpa, a temperature of 400 k and a volume of 50 milliliters. what volume will the gas have at a pressure of 60 kpa and a temperature of 200 kpa?

Answer 1

We use the ideal gas relationship,
PV/T = constant

P₁V₁/T₁ = P₂V₂/T₂

120 x 50  / 400 = 60 x V₂ / 200
V₂ = 50 ml

Answer 2

The new volume of the gas as pressure and temperature are decreased to the given values is 50mL.

What is 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 = 50mL

• Initial pressure P₁ = 120kPa

• Initial temperature T₁ = 400K
• Final pressure P₂ = 60kPa
• Final temperature T₂ = 200K
• Final volume V₂ = ?

P₁V₁/T₁ = P₂V₂/T₂

P₁V₁T₂ = P₂V₂T₁

V₂ = P₁V₁T₂ / P₂T₁

V₂ = ( 120kPa × 50mL × 200K ) / (60kPa × 400K )

V₂ = 1200000 mLkPaK / 24000kPaK

V₂ = 50mL

The new volume of the gas as pressure and temperature are decreased to the given values remains at 50mL.

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

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