Answer: 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 CvΔT=209.4 25RΔT=209.4 ΔT=5R209.4×2 And, Heat exchange=Cm×ΔT where; Cm is molar heat capacity Cm=ΔTHeat Exchange substituting values for Heat Exchange = 418.4 and ΔT=5R209.4×2 Cm=5R Log in to Reply
Answer:
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
CvΔT=209.4
25RΔT=209.4
ΔT=5R209.4×2
And, Heat exchange=Cm×ΔT
where; Cm is molar heat capacity
Cm=ΔTHeat Exchange
substituting values for Heat Exchange = 418.4 and ΔT=5R209.4×2
Cm=5R