A sample occupies 68.0 mL at 40∘C and 95 kPa. What pressure will it have if cooled to −20∘C while the volume remains constant?

Answer 1

P₁ / T₁ = P₂ / T₂

P₁ = 95 kPa

P₂ = ?

T₁ = 40 °C + 273 K = 313 K

T₂ = -20 °C + 273 K = 253 K (**Ms. Lauren incorrectly used positive 20 °C instead of negative 20 °C in her T₂ calculation)

P₂ = P₁ T₂ / T₁ = (95 kPa x 253 K) / 313 K = 77 kPa (rounded to 2 significant figures)

Answer 2

PV = nRT

P1/T1 = P2/T2

40C + 273 = 313K

20C + 273 = 293K

95kPa/313K = P2/293K

P2 = 88.93 K = 89K