What pressure is required to reduce 59 mL of a gas at standard conditions to 15 mL at a temperature of 20◦C? Answer in units of atm.

Question

What pressure is required to reduce 59 mL of a gas at standard conditions to 15 mL at a temperature of 20◦C? Answer in units of atm.

Answer 1

In chemistry, standard temperature and pressure is a temperature of 273.15 K and a pressure of 100 kPa or 0.9869 atm

Use:

where P₁ = 0.9869 atm, V₁ = 59 mL, T₁ = 273.15 K, V₂ = 15 mL, and T₂ = 293.15 K

P₂ = (P₁)•(V₁)•(T₂)/((V₂)•(T₁))

P₂ = (0.9869)(59)(293.15)/((15)(273.15))

P₂ = 17069/4097 = 4.2 atm