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18 December, 10:34

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

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Answers (1)
  1. 18 December, 10:54
    0
    V₂ = 1.41 mL

    Explanation:

    According to general gas equation:

    P₁V₁/T₁ = P₂V₂/T₂

    Given dа ta:

    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
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