Answer to Question #334168 in Chemistry for shimmy

Question

Solve: Helium gas has a volume of 250 mL at 0°C at 1.0 atm. What will be the final

pressure if the volume is reduced to 100 mL at 45°C?

Given: V1 = 250mL                                                       V2 = 100 mL

T₁ = 00C + 273.15= 273.15 K                     T₂ = 450C + 273.15=318.15 K P₁ = 1 atm                                                                    P₂ = ?

Answer 1

Given:

P₁ = 1.0 atm

V₁ = 250 mL

T₁ = 0°C + 273.15 = 273.15 K

P₂ = ???

V₂ = 100 mL

T₂ = 45°C + 273.15 = 318.15 K

Solution:

Combined gas law can be used.

Mathematical expression for the combined gas law:

P₁V₁/T₁ = P₂V₂/T₂

Cross-multiply to clear the fractions:

P₁V₁T₂ = P₂V₂T₁

Divide to isolate P₂:

P₂ = P₁V₁T₂ / V₂T₁

P₂ = (1.0 atm × 250 mL × 318.15 K) / (100 mL × 273.15 K) = 2.9 atm

P₂ = 2.9 atm

Answer: The final pressure will be 2.9 atm