Asample of gas has a volume of 4.28 liters at a temperature of 10.24 degrees celcius and a pressure of 42.20 mmhg. what would be the volume at a pressure of 4.56 kpa and a temperature of 21.3 degrees celcius?

V₂  = 9.14 L

Explanation:

Given data:

Initial volume = 4.28 L

Initial pressure = 42.20 mmHg (42.20/760 = 0.056 atm)

Initial temperature = 10.24 °C (72 +273 = 283.24 K)

Final temperature = 213°C (213+273 = 486 K)

Final pressure = 4.56 KPa (4.56/101.325 = 0.045 atm)

Final volume = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁  P₂

V₂ = 0.056 atm × 4.28 L × 486 K / 283.24 K × 0.045 atm

V₂  =  116.48 atm .L.  K  / 12.75 K . atm

V₂  = 9.14 L

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