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erica [24]
3 years ago
9

A solution is prepared by mixing 2.88 mol of acetone with 1.45 mol of cyclohexane at 30°C. 2nd attempt Calculate the χacetone an

d χcyclohexane in the vapor above the solution. P°acetone = 229.5 torr and P°cyclohexane = 97.6 torr.
Chemistry
1 answer:
Volgvan3 years ago
8 0
First c<span>alculate the mole fraction of each substance:
acetone: 2,88 mol </span>÷ (2,88 mol + 1,45 mol) = 0,665.
cyclohexane: 1,45 ÷ (2,88 mol + 1,45 mol) = 0,335.
Raoult's Law: 
P(total) = P(acetone) · χ(acetone)  + P(cyclohexane) · χ(cyclohexane).
P(total) = 229,5 torr · 0,665 + 97,6 torr · 0,335.
P(total) = 185,3 torr.
χ for acetone: 229,5 torr · 0,665 ÷ 185,3 torr = 0,823.
χ for cyclohexane:  97,6 torr · 0,335 ÷ 185,3 torr = 0,177.

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Explanation:

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In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe

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After that, since the process is isobaric, we can compute the work as:

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V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3

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W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ

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