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The pressure of a gas is reduced from 1200.0 mmHg to 1.11842 atm as the volume of its container is increased by moving a piston from 85.0 mL to 350.0 mL. What would the final temperature be in Celsius if the original temperature was 90.0OC?

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Edufirst

Answer:

          [tex]\large\boxed{T_2=786.\ºC}[/tex]

Explanation:

Ideal gases follow the combined law of gases:

         [tex]P_1V_1/T_1=P_2V_2/T_2[/tex]

Where,

[tex]P_1,V_1, and{\text{ }T_1\text{ are the pressure, temperature, and volume of the gas a state 1}[/tex]

[tex]P_2,V_2, and{\text{ }T_2\text{ are the pressure, temperature, and volume of the gas a state 2}[/tex]

  • Pressure is the absolute pressure and its units may be in any system, as long as they are the same for both states.

  • Also, volume may be in any units, as long as it they are the same for both states.

  • Temperature must be absolute temperature, whose unit is Kelvin.

Your data are:

  • P₁ = 1200.00 mmHg
  • P₂ = 1.11842 atm
  • V₁ = 85.0 mL
  • V₂ = 350.0 mL
  • T₂ = ?
  • T₁ = 90.0ºC

1. Conversion of units:

  • P₁ = 1200.00 mmHg × 1.00000 atm / 760.000 = 1.578947 mmHg
  • T₁ = 90.00ºC + 273.15 = 363.15K

2. Solution

  • Clearing T₂, from the combined gas equation you get:

      [tex]T_2=P_2V_2T_1/(P_1V_1)[/tex]

  • Substituting the data:

         [tex]T_2=1.11842atm\times 350.0ml\times 363.15K/(1.578947atm\times 85.0ml)[/tex]

          [tex]T_2=1,059K[/tex]

  • Convert to celsius:

         [tex]T_2=1059-273.15=786.\ºC[/tex]

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