a 0.175-l flask contains cl2 at 25 °c and 475 mm hg. what is the pressure of cl2 if the volume is increased to 0.800 l and the temperature increased to 55 °c?

Respuesta :

Considering the Combined Law Equation, the pressure of Cl₂ if the volume is increased to 0.800 l and the temperature increased to 55 °C is 114.367 mmHg.

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: if the pressure increases, the volume decreases while if the pressure decreases, the volume increases.

Mathematically, this law is expressed as:

P×V= 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: increasing temperature increases pressure, while decreasing temperature decreases pressure.

Mathematically, this law is expressed as:

P÷T= constant

Charles's law

Charles's law states that the volume is directly proportional to the temperature of the gas when the pressure is constant: if the temperature increases, the volume of the gas increases while if the temperature of the gas decreases, the volume decreases.

Mathematically, this law is expressed as:

V÷T= 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= constant

Considering an initial state 1 and a final state 2:

(P₁×V₁)÷ T₁= (P₂×V₂)÷ T₂

New pressure

In this case, you know:

  • P₁= 475 mmHg
  • V₁= 0.175 L
  • T₁= 25 °C= 298 K (being 0 °C= 273 K)
  • P₂= ?
  • V₂= 0.800 L
  • T₂= 55 °C= 328 K

Replacing in Combined law equation:

(475 mmHg× 0.175 L)÷ 298 K= (P₂× 0.800 L)÷ 328 K

Solving:

[(475 mmHg× 0.175 L)÷ 298 K] × (328 K÷ 0.800 L)= P₂

114.367 mmHg= P₂

Finally, the new pressure will be 114.367 mmHg.

Learn more about Combined Law Equation:

brainly.com/question/4147359

#SPJ1