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If 21 mL f gas is subjected to a temperature change from 10.0C to 120C and a pressure change from 1.0 atm to 15 atm, the new volume is:

1.9 mL
1.0 mL
0.53 mL
440 mL

1 Answer

4 votes

Answer:

V₂ = 1.41 mL

Step-by-step explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 21 mL

Initial pressure = 1 atm

Initial temperature = 10 °C (10 +273 = 283 K)

Final temperature = 12 °C (12 +273 = 285 K)

Final volume = ?

Final pressure = 15 atm

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

V₂ = P₁V₁T₂/T₁ P₂

V₂ = 1 atm × 21 mL × 285 K / 283 K × 15 atm

V₂ = 5985 atm .mL. K / 4245 K. atm

V₂ = 1.41 mL

User Gabriel Glenn
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