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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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Answer:

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Explanation:

The answer choices are:

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<h2>Solution</h2>

Let's see every statement

<u><em>A.) Gases are readily soluble in water at higher temperatures.</em></u>

This is false. At higher temperature, the gas molecules will have higher kinetic energy, meaning that they will move faster and escape more easily from the liquid phase. Thus, gases are less soluble at higher temperatures.

<u><em>B.) Carbon dioxide gas will be less soluble in water when the partial pressure is low.</em></u>

This is TRUE.

HIgher pressures "push" the gas molecules into the liquid phase. When more molecules get dissolved the pressure is partially relieved.

This relation is stated by Henry's Law: the solubility of gases in liquids is directly proportional to the pressure of the gases above the surface of the solution. Thus, carbon dioxide will me less soluble in water when the pressure is low.

<u><em>C.) Zinc sulfate is soluble in water.</em></u>

This is TRUE.

You must resort to the rules of solubility or search the properties of this salt.

Th solubility rules state that most sulfates are soluble in water. Exceptions are sulfates of Ag⁺,Ca²⁺, Sr²⁺, Ba²⁺, Hg²⁺, and Pb²⁺. Thus, you can expect that zinc sulfate is soluble in water.

<u><em>D.) Propanol and heptane are miscible.</em></u>

This is false.

Propanol is a polar compound. It is miscible with water because water is polar too. Heptane is not polar. Thus, following the rule that "like dissolves like" propanol and heptane are not miscible.

<u><em></em></u>

<u><em>E.) Table salt dissolves in water to form a solution.</em></u>

This is TRUE.

This is a classical example of an ionic compound (sodium chloride) that is soluble in water (a polar solvent), showing the rule that polar solvents dissolve polar and ionic compounds.

It is also an everyday fact: rice, for instance, absorbs salt from water because the salt is dissolved in the water.

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Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
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Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

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