If an 85.0 mL container of helium gas at standard pressure is heated from 20 C to 91 C and the pressure is increased to 2.8 atm then what would the new volume be for the He gas?

V₂ = 37.71 mL

Explanation:

Given data:

Initial volume = 85.0 mL

Initial pressure = 1 atm

Initial temperature = 20°C (20 +273 = 293 K)

Final temperature = 91°C (91+273 = 364 K)

Final volume = ?

Final pressure = 2.8 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:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1 atm× 85.0 mL × 364 K / 293 K × 2.8 atm

V₂ = 30940 atm.mL.K / 820.4 K.atm

V₂ = 37.71 mL

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

the answer is 0.8m.

by: nicholas