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vivado [14]
3 years ago
15

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 compressi

on. 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.
Chemistry
1 answer:
oksian1 [2.3K]3 years ago
8 0

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

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The molecular formula of the given compound is $\mathrm{H}_{2} \mathrm{CO}_{3}$ also known as  Carbonic acid.

<h3>What is empirical formula and molecular formula?</h3>

The simplest whole-number ratio of the various atoms in a compound is represented by an empirical formula.

The precise number of various atom types present in a compound's molecule is indicated by the molecular formula.

Given that,

H = 3.3%

C = 19.3%

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Therefore, the ratio of the atoms of C, H and O = 3.3 : 1.60 : 4.83

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Let the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) \mathrm{n}$

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