Ask question

Ask question

All categories

3 years ago

6

Find the fugacity coefficient of a gaseous species of mole fraction 0.4 and fugacity 25 psia in a mixture at a total pressure of

50 psia

1 answer:

Oksana_A [137]3 years ago

8 0

<u>Answer:</u> The fugacity coefficient of a gaseous species is 1.25

<u>Explanation:</u>

Fugacity coefficient is defined as the ratio of fugacity and the partial pressure of the gas. It is expressed as $\bar{\phi}$

Mathematically,

$\bar{\phi}_i=\frac{\bar{f_i}}{p_i}$

Partial pressure of the gas is expressed as:

$p_i=\chi_iP$

Putting this expression is above equation, we get:

$\bar{\phi}_i=\frac{\bar{f_i}}{\chi_iP}$

where,

$\bar{\phi}_i$ = fugacity coefficient of the gas

$\bar{f_i}$ = fugacity of the gas = 25 psia

$\chi_i$ = mole fraction of the gas = 0.4

P = total pressure = 50 psia

Putting values in above equation, we get:

$\bar{\phi}_i=\frac{25}{0.4\times 50}\\\\\bar{\phi}_i=1.25$

Hence, the fugacity coefficient of a gaseous species is 1.25

You might be interested in

Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) A reaction mixture is made containing an initial [SO2Cl2] of 2.2×10−2M

Andrej [43]

The value of equilibrium constant is equal to the quotient of the products raised to its stoichiometric coefficient over the reaction's reactants raised to its respective stoichiometric coeff. The equation is Kc=[SO2][Cl2]/[SO2Cl2]= [1.3*10^-2][1.3*10^-2]/[2.2*10^-2-<span>1.3*10^-2]=0.0188. The final answer is Kc=0.0188.</span>

6 0

3 years ago

Plants can provide the materials that animals use in cellular respiration, and animals can provide some of the materials that pl

IrinaK [193]

Answer:

it produce atp

Explanation:

.............

6 0

3 years ago

The smell of fresh cut pine is due in part to the cyclic alkene called pinene. Given the following data of pinene: Vapor pressur

Sveta_85 [38]

Answer: Heat of vaporization is 41094 Joules

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

$ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = initial pressure at 429 K = 760 torr

$P_2$ = final pressure at 415 K = 515 torr

= enthalpy of vaporisation = ?

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 429 K

$T_2$ = final temperature = 515 K

Now put all the given values in this formula, we get

$\log (\frac{515}{760}=\frac{\Delta H}{2.303\times 8.314J/mole.K}[\frac{1}{429K}-\frac{1}{415K}]$

$\Delta H=41094J$

Thus the heat of vaporization is 41094 Joules

4 0

3 years ago

Blank is radiant energy emitted from the sun.

Bess [88]

None, it’s solar radiation

7 0

3 years ago

Read 2 more answers

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

3 years ago