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A 988ml sample of air is at 852 mm Hg and 34.1 C. What will the temperature of this gas be, in Fahrenheit, at 955 mm Hg and a volume of 602 mL

1 Answer

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Answer:

T₂ = 63.78 °F

Step-by-step explanation:

Given data:

Initial Volume = 988 mL

Initial Pressure = 852 mmHg

Initial temperature = 34.1 °C

Final pressure = 955 mmHg

Final volume = 602 mL

Final temperature = ?

Solution:

(34.1 °C × 9/5) +32 = 93.38 °F

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂

T₂ = P₂V₂T₁/P₁V₁

T₂ = 955 mmHg × 602 mL × 93.38 °F / 852 mmHg× 988mL.

T₂ = 53685095.8 °F /841776

T₂ = 63.78 °F

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