Respuesta :
Answer:
Cp = 5.982 R
∴ R: ideal gas constant
Explanation:
expand reversibly and adiabatically:
∴ T1 = 298.15 K
∴ T2 = 248.44 K
∴ P1 = 1522.2 Torr
∴ P2 = 613.85 Torr
⇒ δU = δQ + δW......first law
∴ Q = 0....adiabatically
⇒ δU = CvδT = δW = - PδV
⇒ CvδT = - nRT/V δV
⇒ CvδT/nT = - R δV/V
∴ Cv/n = Cv,m
⇒ Cv,m Ln(T2/T1) = R Ln(V1/V2)
⇒ Cv,m ( - 0.1823 ) = R ( - 0.9082 )
⇒ Cv.m = 4.982 R
∴ Cp,m - Cv,m = R...."perfect" gas
⇒ Cp,m = R + Cv,m
⇒ Cp,m = R + 4.982 R
⇒ Cp,m = 5.982 R
∴ Cp,m = Cp/n
assuming: n = 1 mol fluorocarbon gas
⇒ Cp = 5.982 R
∴ R: ideal gas constant
Answer:
Cp=42.58 J/K*mol
Explanation:
To solve this exercise we will use the knowledge and formulas of the reversible adiabatic expansion to calculate the value of Cp. In the attached image the answer given is calculated step by step.
