The formula V1/T1 = V2/T2 applies to ____

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

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Gaseous dichlorine monoxide decomposes readily to chlorine (green) and oxygen (red) gases.

Stolb23 [73]

The reactants are oxygen and nitrogen monoxide.

<h3>What is nitrogen monoxide?</h3>

Nitrogen oxide, also known as nitrogen monoxide or nitric oxide, is an inert gas with the chemical formula NO. It is one of the main nitrogen oxides. Free radical nitric oxide (•N=O or •NO) possesses an unpaired electron, which is commonly indicated by a dot in its chemical formula. As a heteronuclear diatomic molecule, nitric oxide also contributed to the development of early modern theories of chemical bonding.

Nitric oxide is a chemical compound that occurs in combustion systems and can be produced by lightning during thunderstorms. It is a crucial intermediate in industrial chemistry. In many physiological and pathological processes in animals, including humans, nitric oxide serves as a signaling molecule.

To learn more about nitrogen monoxide from the given link:

brainly.com/question/13428103

#SPJ4

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A piece of limestone erodes due to acid rain. This process can be best described as a...

scZoUnD [109]

A. Slow Chemical Change

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Given the chemical reaction: Hg2+(aq) + Cu(s) + Hg(s) + Cu2+(aq)

Pavel [41]

Answer: When the reaction reaches equilibrium, the cell potential will be 0.00 V

Explanation:

Equilibrium state is the state when reactants and products are present but the concentrations does not change with time.

The equilibrium is dynamic in nature and the reactions are continuous in nature. Rate of forward reaction is equal to the rate of backward reaction.

The standard emf of a cell is related to Gibbs free energy by following relation:

$\Delta G=-nFE^0$

The Gibbs free energy is related to equilibrium constant by following relation:

$\Delta G=-2.303RTlog K$

For equilibrium $\Delta G=0$

Thus $0=-nFE^0$

$E^0=0$

Thus When the reaction reaches equilibrium, the cell potential will be 0.00 V

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3 years ago

Which atom gives up its electrons most easily?<br><br> THE ANSWER IS “Rb”

Sedbober [7]

Answer:

cesium

In particular, cesium (Cs) can give up its valence electron more easily than can lithium (Li). In fact, for the alkali metals (the elements in Group 1), the ease of giving up an electron varies as follows: Cs > Rb > K > Na > Li with Cs the most likely, and Li the least likely, to lose an electron

Explanation:

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The formula V1/T1 = V2/T2 applies to _____.