A3.2 liter sample of a gas is at 40 oc and 1.0 atmosphere of pressure. if the temperature decreases to 20 oc and the pressure decreases to 0.60 atmospheres, what is the new volume of the gas in liters?

V₂  = 5 L

Explanation

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 3.2 L

Initial pressure = 1 atm

Initial temperature = 40 °C (40 +273 = 313 K)

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

Final pressure = 0.60 atm

Final volume = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 1 atm × 3.2 L × 293 K / 313 K × 0.60 atm

V₂  =  937.6 atm .L.  K  / 187.8 K . atm

V₂  = 5 L

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.