Question

One mole of an ideal gas, for which CV,m = 3/2R, initially at 298 K and 1.00 × 105 Pa undergoes a reversible adiabatic compression. At the end of the process, the pressure is 1.00 × 106 Pa. Calculate the final temperature of the gas. Calculate q, w, ΔU, and ΔH for this process. Show all equation derivations as necessary. Given equations include but not limited to: pv=nRt, deltaU=q+w, q=ncdeltaT.

Answer 1

Answer:

• final temperature (T2) = 748.66 K
• ΔU = w = 5620.26 J
• ΔH = 9367.047 J
• q = 0

Explanation:

ideal gas:

• PV = RTn

reversible adiabatic compression:

• δU = δq + δw = CvδT

∴ q = 0

∴ w = - PδV

⇒ δU = δw

⇒ CvδT = - PδV

ideal gas:

⇒ PδV + VδP = RδT

⇒ PδV = RδT - VδP = - CvδT

⇒ RδT - RTn/PδP = - CvδT

⇒ (R + Cv,m)∫δT/T = R∫δP/P

⇒ [(R + Cv,m)/R] Ln (T2/T1) = Ln (P2/P1) = Ln (1 E6/1 E5) = 2.303

∴ (R + Cv,m)/R = (R + (3/2)R)/R = 5/2R/R = 2.5

⇒ Ln(T2/T1) = 2.303 / 2.5 = 0.9212

⇒ T2/T1 = 2.512

∴ T1 = 298 K

⇒ T2 = (298 K)×(2.512)

⇒ T2 = 748.66 K

⇒ ΔU = Cv,mΔT

⇒ ΔU = (3/2)R(748.66 - 298)

∴ R = 8.314 J/K.mol

⇒ ΔU = 5620.26 J

⇒ w = 5620.26 J

• H = U + nRT

⇒ ΔH = ΔU + nRΔT

⇒ ΔH = 5620.26 J + (1 mol)(8.314 J/K.mol)(450.66 K)

⇒ ΔH = 5620.26 J + 3746.787 J

⇒ ΔH = 9367.047 J