The structural formula of 2-methyl-propane<br>-2-ol

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

2 years ago

What mass of KNO, will dissolve in 100 g of water at 100°C?

Sphinxa [80]

Answer:

About 170-180 grams of potassium nitrate are completely dissolved in 100 g.

Explanation:

Hello!

In this case, according to the reported solubility data for potassium nitrate at different temperatures on the attached picture, it is possible to bear out that about 170-180 grams of potassium nitrate are completely dissolved in 100 g; considering that the solubility is the maximum amount of a solute that can be dissolved in a solvent, in this case water.

Best regards!

3 0

2 years ago

A)The lower the greenhouse gases, the higher the ocean temperature.

HACTEHA [7]

The higher the greenhouse gases, the higher the ocean temperature

"The ocean absorbs most of the excess heat from greenhouse gas emissions, leading to rising ocean temperatures"

hope this helps :)

6 0

2 years ago

Read 2 more answers

Porque una mezcla tiene composición variable?

LuckyWell [14K]

¿ cuál es la pregunta que intentas hacer?

7 0

3 years ago

2. if 0.20 m fe3 had been used instead of 0.020 m fe3 , how would the numerical value of the rate constant and the activation en

dezoksy [38]

the calculated value is Ea is 18.2 KJ and A is 12.27.

According to the exponential part in the Arrhenius equation, a reaction's rate constant rises exponentially as the activation energy falls. The rate also grows exponentially because the rate of a reaction is precisely proportional to its rate constant.

At 500K, K=0.02s−1

At 700K, k=0.07s −1

The Arrhenius equation can be used to calculate Ea and A.

RT=k=Ae Ea

lnk=lnA+(RT−Ea)

At 500 K,

ln0.02=lnA+500R−Ea

500R Ea (1) At 700K lnA=ln (0.02) + 500R

lnA = ln (0.07) + 700REa (2)

Adding (1) to (2)

700REa100R1[5Ea-7Ea] = 0.02) +500REa=0.07) +700REa.

=ln [0.02/0 .07]

Ea= 2/35×100×8.314×1.2528

Ea =18227.6J

Ea =18.2KJ

Changing the value of E an in (1),

lnA=0.02) + 500×8.314/18227.6

= (−3.9120) +4.3848

lnA=0.4728

logA=1.0889

A=antilog (1.0889)

A=12.27

Consequently, Ea is 18.2 KJ and A is 12.27.

Learn more about Arrhenius equation here-

brainly.com/question/12907018

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5 0

1 year ago

The structural formula of 2-methyl-propane-2-ol​

The structural formula of 2-methyl-propane-2-ol