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.

Question

Skipper

Chemistry

7 April, 03:32

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.

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Judah Richardson · Answer 1

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