A gas goes from 1L at 273 K to 2 L when heated. What is the new temp?
Given : Diatomic molecule at 273K
‘q’ absorbed = positive = +100Cal = 100 x 4.184J = 418.4J
‘W’ done by system = negative = -209J
By first law of thermodynamics;
ΔU = q + W = 418.4 + (-209) = 209.4J
We know for diatomic molecule Cv=25R and CvΔT=ΔU
And, Heat exchange=Cm×ΔT
where; Cm is molar heat capacity
substituting values for Heat Exchange = 418.4 and ΔT=5R209.4×2
You must be logged in to post a comment.