Question

Two thermally insulated vessels are connected by a narrow tube fitted with a valve that is initially closed as shown in the figure below. One vessel of volume V1 = 17.1 L, contains oxygen at a temperature of T1 = 310 K and a pressure of P1 = 1.80 atm. The other vessel of volume V2 = 23.0 L contains oxygen at a temperature of T2 = 470 K and a pressure of P2 = 1.80 atm. When the valve is opened, the gases in the two vessels mix and the temperature and pressure become uniform throughout. (a) What is the final temperature? Incorrect: Your answer is incorrect. K (b) What is the final pressure? atm

Answer 1

a)

P₁ = pressure in vessel 1 = 1.80 atm = 1.80 x 1.01 x 10⁵ Pa

V₁ = Volume in Vessel 1 = 17.1 L = 17.1 x 10^-3 m³?

T₁ = 310 K

n₁ = number of moles in vessel 1 = P₁ V₁/(R T₁) = (1.80 x 1.01 x 10⁵) (17.1 x 10^-3 )/((8.314) (310)) = 1.21

P₂ = pressure in vessel 2 = 1.80 atm = 1.80 x 1.01 x 10⁵ Pa

V₂ = Volume in Vessel 2 = 23 L = 23 x 10^-3 m³?

T₂ = 470 K

n₂ = number of moles in vessel 2 = P₁ V₁/(R T₁) = (1.80 x 1.01 x 10⁵) (23 x 10^-3 )/((8.314) (470)) = 1.07

let the equilibrium temperature be "T"

using conservation of energy

n₁ (T - 310) = n₂ (470 - T)

(1.21)(T - 310) = (1.07) (470 - T)

T = 385.01 K

b)

n = n₁ + n₂ = 1.21 + 1.07 = 2.28

V = V₁ + V₂ = 17.1 x 10^-3 + 23 x 10^-3 = 40.1 x 10^-3 m³

Using the equation

PV = n R T

P (40.1 x 10^-3) = (2.28)(8.314) (385.01)

P = 1.82 x 10⁵ Pa = 1.802 atm