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Mars2501 [29]
3 years ago
14

Fill in the blanks with the words given below-

Chemistry
1 answer:
ioda3 years ago
8 0

Answer:

1. Metal.

2. Atom.

3. Homogeneous

4. Compounds.

5. Lustrous

6. Saturated.

7. Colloidal; true.

8. Homogeneous.

Explanation:

1. An element which are sonorous are called metal.

2. An element is made up of only one kind of atom.

3. Alloys are homogeneous mixtures.

4. Elements chemically combines in fixed proportion to form compounds.

5. Metals are lustrous and can be polished.

6. A solution in which no more solute can be dissolved is called a saturated solution.

7. Milk is a colloidal solution but vinegar is a true solution.

8. A solution is a homogeneous mixture.

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Answer:

a). P = 688 atm

b). P = 1083.04 atm

c).Δ G = 16.188 J/mol    

Explanation:

a). Fugacity 'f' can be calculated from the following equations :

$\ln \frac{f}{P}=\int^P_0\left(\frac{Z-1}{P}\right) dP$

where, P = pressure ,  Z = compressibility

Now, the virial equation is :

$PV=R(1+6.4\times 10^{-4} P)$  ........(1)

Also, PV=ZRT for real gases .......(2)

∴ $ZRT=RT(1+6.4\times 10^{-4} P)$

  $Z=1+6.4\times 10^{-4} P$

So from the fugacity equation ,

$\ln \frac{f}{P}=\int^P_0\left(\frac{1+6.4\times 10^{-4}P-1}{P}\right) dP$

$\ln \frac{f}{P}=\int^P_0\left(\frac{6.4\times 10^{-4}P}{P}\right) dP$

$\ln \frac{f}{P} = 6.4 \times 10^{-4} P$

$f=Pe^{6.4 \times 10^{-4} P}$

Putting the value of P = 500 atm in the above equation, we get,

f = 688 atm

b). Given f = 2P

    $2P=PE^{6.4 \times 10^{-4}P}$

   $2=E^{6.4 \times 10^{-4}P}$

  $\ln 2 = 6.4 \times 10^{-4}P$

  ∴  P = 1083.04 atm

c). dG = Vdp -S dt  at constant temperature, dT = 0

Therefore, dG = V dp

$\int^{G_2}_{G_1}dG =\int^{P_2}_{P_1}V dp $

            $=\int^{P_2}_{P_1}\frac{RT}{P}\left(1+6.4 \times 10^{-4}P\right) dP$

           $=\int^{P_2}_{P_1}RT\frac{dp}{P}+\int^{P_2}_{P_1}RT6.4 \times 10^{-4} dP$

$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$

$\Delta G=8.314\times 298[\ln\frac{500}{1}+6.4 \times 10^{-4}(500-1)]$

Δ G = 16.188 J/mol  

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