Question

The constant-volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabatically and reversibly. If a decrease in pressure is also measured, we can use it to infer the value of γ (the ratio of heat capacities, Cp/CV) and hence, by combining the two values, deduce the constant pressure heat capacity. A fluorocarbon gas was allowed to expand reversibly and adiabatically to twice its volume; as a result, the temperature fell from 298.15 K to 248.44 K and its pressure fell from 1522.2 Torr to 613.85 Torr. Evaluate Cp.

Answer 1

Cp=42.58 J/K*mol

Step-by-step 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.

Answer 2

Cp = 5.982 R

∴ R: ideal gas constant

Step-by-step 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