a sealed cylinder fitted with a movable piston contains ideal gas at 27°c, pressure 0.500 × 105 pa, and volume 1.1 m3. what will be the final temperature if the gas is compressed to 0.800 m3 and the pressure rises to 0.820 × 105 pa?

Respuesta :

Considering the combined law equation, the final temperature if the gas is compressed to 0.800 m³ and the pressure rises to 0.820×10⁵ Pa is 357.82 K.

Boyle's law

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant.

Mathematically, Boyle's law states that the product of pressure and volume always has the same value:

P×V= k

where:

  • P is the pressure.
  • V is the volume.
  • k is a constant.

Charles' law

Charles' law states that the volume is directly proportional to the temperature of the gas when the pressure is constant.

Mathematically, Charles' law states that the ratio between volume and temperature always has the same value:

V÷T=k

where:

  • V is the volume.
  • T is the temperature.
  • k is a constant.

Gay-Lussac's law

Gay-Lussac's law states that the pressure of a gas is directly proportional to its temperature when the volume is constant.

Mathematically, Charles' law states that the ratio between pressure and temperature always has the same value:

P÷T=k

where:

  • P is the pressure.
  • T is the temperature.
  • k is a constant.

Combined law equation

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

(P×V)÷T= k

where:

  • P is the pressure.
  • V is the volume.
  • T is the temperature.
  • k is a constant.

Analyzing an initial state 1 and a final state 2, it is fulfilled:

(P₁×V₁)÷T₁= (P₂×V₂)÷T₂

Final temperature in this case

In this case, you know:

  • P₁= 0.500×10⁵ Pa
  • V₁= 1.1 m³
  • T₁= 27 °C= 300 K (being 0 °C= 273 K)
  • P₂= 0.820×10⁵ Pa
  • V₂= 0.800 m³
  • T₂= ?

Replacing in the combined gas law:

(0.500×10⁵ Pa× 1.1 m³)÷ 300 K= (0.820×10⁵ Pa× 0.800 m³)÷T₂

Solving:

[(0.500×10⁵ Pa× 1.1 m³)÷ 300 K]× T₂= (0.820×10⁵ Pa× 0.800 m³)

T₂= (0.820×10⁵ Pa× 0.800 m³)÷ [(0.500×10⁵ Pa× 1.1 m³)÷ 300 K]

T₂= 357.82 K

Finally, the final temperature is 357.82 K.

Learn more about combined law equation:

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