Question

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?

Answer 1

Answer:

          $\large\boxed{T_2=786.\ºC}$

Explanation:

Ideal gases follow the combined law of gases:

         $P_1V_1/T_1=P_2V_2/T_2$

Where,

$P_1,V_1, and{\text{ }T_1\text{ are the pressure, temperature, and volume of the gas a state 1}$

$P_2,V_2, and{\text{ }T_2\text{ are the pressure, temperature, and volume of the gas a state 2}$

• 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:

      $T_2=P_2V_2T_1/(P_1V_1)$

• Substituting the data:

         $T_2=1.11842atm\times 350.0ml\times 363.15K/(1.578947atm\times 85.0ml)$

          $T_2=1,059K$

• Convert to celsius:

         $T_2=1059-273.15=786.\ºC$