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A 0.617-mol sample of CO_2(g) initially at 298 K a

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Answer #1

q = nCpdT

n = no of moles of CO2 gas = 0.617 mol

Cp = constant pressure heat capacity of CO2 = 7R/2 = (7/2) x 8.314 J/K/mol

dT = 13.1 K

Hence,

q = nCpdT

= 0.617 mol x (7/2) x 8.314 J/K/mol x 13.1 K

= 235.2 J

Therefore, q = 235.2 J

----------------------------------------------------------

We know that Cp-Cv= R

Cv = Cp-R = (7/2)R -R = (5/2)R

We know that

  ΔU = nCvdT

= nx(5/2)Rx dT

= 0.617 mol x (5/2) x 8.314 J/K/mol x 13.1 K

= 168 J

Therefore,   ΔU = 168 J

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