What is the final temperature in ∘c if the process is isothermal?

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

6K + B2O3 → 3K2O + 2B

atroni [7]

Answer:

104.84 moles

Explanation:

Given data:

Moles of Boron produced = ?

Mass of B₂O₃ = 3650 g

Solution:

Chemical equation:

6K + B₂O₃ → 3K₂O + 2B

Number of moles of B₂O₃:

Number of moles = mass/ molar mass

Number of moles = 3650 g/ 69.63 g/mol

Number of moles = 52.42 mol

Now we will compare the moles of B₂O₃ with B from balance chemical equation:

B₂O₃ : B

1 : 2

52.42 : 2×52.42 = 104.84

Thus from 3650 g of B₂O₃ 104.84 moles of boron will produced.

6 0

2 years ago

Which pair of statements below is correct? Multiple Choice An octet is formed via ionic bonding when one or more valence electro

tresset_1 [31]

Answer:

An octet is formed via ionic bonding when one or more valence electrons are transferred from one atom to another.

An octet is formed via covalent bonding when valence electrons are shared between atoms.

An octet is always formed via ionic bonding

Explanation: The essence of bonding is stability. An octet or duplet state is formed when one or more valence electrons are shared. when the electrons are shared, the type of bond formed is a covalent bond. An octet is formed via ionic bonding when one or more valence electrons are transferred from one atom to another.

3 0

2 years ago

Read 2 more answers

Which sentence is a scientific statement? A. There is a substance around us that cannot be detected. B. Kids should go to school

stellarik [79]

Answer:

Experimentation is the best way to obtain knowledge.

Explanation:

This is a statement and it is scientific

6 0

2 years ago

The Kp for the reaction below is 1.49 × 108 at 100.0°C:

zhenek [66]

In an equilibrium mixture of the three gases, PCO = PCl2 = 2.22 × 10-4 atm. The partial pressure of the product, phosgene (COCl2), is kp=(COCl2)/(CO)(Cl2) which is $1.48*10^=x/(2.24*10-4)^2$ . So, the correct answer is <span>A) 7.34.</span>

8 0

2 years ago

Read 2 more answers