Answer:
a. T₂ = 901.43 ° K
b. T₂ = 843.85 ° K
Explanation:
Given
η = 0.165 mol , Q = 690 J , V = 280 cm³ , P = 3.40 x 10 ⁶ Pa , T₁ = 700 ° K
Using the equation that describe the experiment of heat
Q = η * Cv * ΔT
a.
Nitrogen Cv = 20.76 J / mol ° K
ΔT = T₂ - T₁
T₂ = [ Q / ( η * Cv) ] + T₁
T₂ = [ 690 J / ( 0.165 mol * 20.76 J / mol ° K ) ] + 700 ° K
T₂ = 901.43 ° K
b.
Nitrogen Cp = 29.07 J / mol ° K
T₂ = [ Q / ( η * Cv) ] + T₁
T₂ = [ 690 J / ( 0.165 mol * 29.07 J / mol ° K ) ] + 700 ° K
T₂ = 843.85 ° K
1.6 is the answer you are looking for.
