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SOLVED

CSIR UGC JRF NET

CHEMICAL SCIENCES

PAPER 1 (PART-B)

SERIES-3

NOTE: Related and additional questions appeared in previous GATE exams are also

solved.

61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

61) The polymeric species (SN)nis a / an

1. three dimensional conductor

2. two dimensional conductor

3. insulator

4. one dimensional conductor

Explanation:

* (SN)n, known as polythiazyl, is a linear polymeric sulfur nitride. It is a one dimensional

conductor. It was found to be a superconductor at very low temperatures (below 0.26 K).

* It is used as barrier electrode in ZnS junctions.

* It increases the quantum efficiency of blue emission by a factor of 100 compared to gold.

* It increases the efficiency of GaAssolar cells by up to 35%.

Additional information:

Structure of (SN)n

N

SN

S N

SN

S N

SN

S

1200 1060

It is a linear polymer (n = up to 2000). It exists in several resonance forms.

Preparation:

Polythiazyl is synthesized by passing Tetrasulfur tetranitride (S4N

4) over silver metal. In this

conversion, silver is sulfided to silver sulfide which catalyses the conversion of S4N

4to

Disulfur dinitride (S2N

2), which readily polymerized to (SN)

n.

S4

N4

+ 8AgAg

2S + 2N

2

S4N4Ag2S (as catalyst)

77 K2S2N2

2S2N2

(SN)n1) sublimes to surface at 0oC

2) then undergoes thermal polymerization

Additional questions:

61.1) Draw the structure of S4N

4.

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Ans:- S

S

S

N N

NN

S

61.2) Write the equations for preparation of S4N

4from SCl

2.

Ans:- S4

N4

was prepared by the reaction of ammonia with SCl2

in carbon tetrachloride followed

by extraction into dioxane.

24 SCl2+ 64 NH

3--------------> 4 S

4N

4+ S

8+ 48 NH

4Cl

61.3) What is the difference between cyclophosphazanes and cyclophosphazenes?

Ans:- Cyclophosphazanes: saturated cyclic systems involving P(III)nitrogen linkages; mostly

four membered.

Cyclophosphazenes: unsaturated rings with P(V)nitrogen bonds; mostly six or eight

membered.

61.4) Write the equation for the synthesis of (Cl2NP)

2?

Ans:- Reaction of phosphorus pentachloride with ammonium chloride in a high-boiling solvent,

like 1,1,2,2-tetrachloroethane yields (Cl2PN)

2.

TO BE COMPLETED

Practice questions:

1) The structure of P4N

4Cl

8is puckered whereas that of P

4N

4F

8is planar because:

a) F is more electronegative than Cl

b) F is smaller in size than that of Clc) F is more polarizable than Cl

d) Extent of -electron delocalization more in P4N

4Cl

8than in P

4N

4F

8

63) The molar absorptivity at max is minimum for

1. [Mn(H2O)

6]2+

2. [Cr(H2O)

6]2+

3. [Co(H2O)

6]2+

4. [Fe(H2O)

6]2+

Explanation:

* The molar absorptivity or extinction coefficient () indicate

s the intensity of absorption of

light radiation during the excitation of electrons. If this value is large then the the complex

have intense color and otherwise it will have pale color.

* The intensity can be determined from the following quantum mechanical selection rules that

state whether the transitions are allowed (intense color) or not allowed (pale color)

a) Symmetry forbidden or Laporte forbidden ( = 1l ): If the molecule has centre ofsymmetry, the transitions from one centrosymmetric orbital (that with centre of inversion) to

another are forbidden. i.e., transitions within a given set of p or d orbitals (i.e. those which

only involve a redistribution of electrons within a given subshell) are forbidden.

In other words, g-->g or u-->u transitions are not allowed.

b) Spin forbidden (S = 0 ): The number of unpaired electrons cannot change upon

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excitation i.e., no electron spin-flip is allowed. Always transition between triplet to triplet and

singlet to singlet states are allowed.

ANSWER TO THE QUESTION:

* All the molecules are octahedral with centre of symmetry and hence the excitations are

symmetry forbidden. (In case of Cr2+there is Jahn-Teller distortion)

* Check for whether spin forbidden or not?

The electron distributions are shown below (remember H2O is a weak field ligand)

Mn2+

- 3d5

or t2g3

eg2

spin forbidden

Cr2+ - 3d4 or t2g

3eg

1 one spin allowed transition

Co2+ - 3d7 or t2g

5eg

2 three spin allowed transitions

Fe2+ - 3d6 or t2g

4eg

2 one spin allowed transition

* As the transitions in Mn2+are both symmetry and spin forbidden, the molar absorptivity is

minimum for first complex, [Mn(H2O)6]2+

. Hence it is pale pink in color.

Note: Extinction coefficients for tetrahedral complexes are expected to be around 50-100

times larger than for octrahedral complexes. Why? Ans: Do not possess centre of inversion.

Additional questions

63.1) KMnO4shows an intense pink colour, while KReO

4is colourless.Explain.

Ans:- KMnO4shows an intense pink colour, while KReO

4is colourless.

Explanation: Mn7+in MnO4- has d0configuration. Hence no d-d transition. Still it shows colordue to ligand to metal charge transfer (LMCT) phenomenon. MnO

4

-absorbs yellow and

transmits violet part of light.

Charge transfer spectra:Charge transfer (CT) is of two types viz.,

i) Ligand to Metal Charge Transfer (LMCT)

ii) Metal to Ligand Charge Transfer (MLCT)

* Remember! The energy required for CT is greater than the energy for d-d transitions. But it

is spin and symmetry allowed and hence intense absorptions are possible. i.e., extinctioncoefficients are higher.

LMCT: In LMCT, an electron is transferred from ligand to the

metal, which is therefore

reduced in the excited state. Charge transfer is analogous to an internal redox reaction, and

hence absorption energies can be correlated with trends in redox properties. The more posi-

tive the redox potential concerned, the easier such reduction will be, and so the lower the

LMCT energy. LMCT transitions in the visible region of the spectrum give intense color as in

case of permanganate ion (MnO4

-).

The general LMCT energy trends in some d0species are:i) LMCT energy decreases towards the right in the 3d series.

e.g. VO4

3-> CrO4

2-> MnO4

-

ii) LMCT energy increases down the transition metal group.

e.g. MnO4

-< TcO4

-< ReO4

-

The above orders of LMCT energy are reflected in the changing colors of the ions and as

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the transition moves progressively to higher energy out of the visible spectrum into the UV.

e.g.

i) MnO4

- deep purple (absorbs yellow color (low energy) and transmits violet)

ii) CrO4

2- deep yellow (absorbs violet (high energy) and transmits yellow)

iii) VO4

3- pale yellow (as in above case)

Now the answer to final part of the question. As the energy required for the charge trans-

fer in case of ReO4

-is in the UV region, it does not show any color. It absorbs UV and trans-

mits entire VIBGYOR (= white).

Other examples:

i) CdS: The color of artists pigment cadmium yellow is due to transition from S2-(p

orbital) to Cd2+(empty 5s orbital).

ii) HgS: It is red due to S2-(p) to Hg2+ (6s) transition.

Note: In Cd2+and Hg2+, d-d transitions are not possible due to d10configuration.

MLCT:The less common, MLCT results in the reduction of the metal center. The charge

transfer occurs from metal to ligand with emty orbital especially low lying * orbital.e.g. Tris(2,2-bipyridyl)ruthenium(II) - [Ru(bpy)

3]2+

[Fe(bpy)3]2+

[Fe(phen)3]2+ -- ferroin

Note: phen = 1,10-phenanthroline

Optical spectroscopy (UV-Visible) is a powerful technique to assign and characterize

charge transfer bands in these complexes.

Additional questions:63.1) In general, UV-Visible absorption peaks of transition metal ions are broad, whereas

those of f-block metal ions are sharp. Explain?

Ans:- In transition metals, absorption peaks appear due to d-d transitions, which are affected by

the external environment (like solvent and ligand effects). Hence absorptions are broad.

In case of f-block elements, the f orbitals are deep inside the atom and are screened from

the external environment. Hence the peaks are sharp. However, the absorptions are less

intense.

63.2) Explain why ff transitions of lanthanide complexes are weaker than dd transitionsof transition metal complexes in the electronic spectra?

Ans:- * Both dd and ff transitions are parity forbidden (symmetry f

orbidden).

* But in the case of dd spectra, vibronic coupling lowers the symmetry round the metal ion,

so d/p mixing can occur and the dd transitions are to some extent allowed. (It is the reason

why the peaks are observed even though they are forbidden)

* Because the 4f orbitals are so contracted (as they are deep inside) compared with d orbitals,

f/p mixing and vibronic coupling are much more limited, thus the ff transitions are less

allowed and so the peaks are less intense.

Note: The centre of symmetry of complexes is momentarily destroyed due to asymmetric

molecular vibrations. This is called vibronic coupling.

63.3) Fe(SCN)2+is blood red in color eventhough d-d transitions in Fe3+ion with d5configu-

ration are forbidden. Explain.

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Ans:- Fe(SCN)2+, correctly represented as [Fe(NCS)(H2O)

5]2+, is an octahedral complex. The

configuration of Fe3+is t2g

3eg

2. Hence the d-d transitions in them are spin forbidden (though

Laporte allowed).

But the intense red color is due to charge transfer from SCN-to Fe(III) ion (LMCT).

63.4) Predict the line width and the intensities of peaks in the electronic spectra of CuCl4

2-

and Eu(H2O)

8

3+complexes.

Ans:- The peaks are broad and intense in case of CuCl4

2-.

* Intense due to LMCT possible from Cl-

ligand to Cu(II).* The d-d component is also strong as the d-d transitions are Laporte allowed as the complex

is tetrahedral (no inversion symmetry).

* Broad because d-d transitions are vibronically coupled.

In case of Eu(H2O)

8

3+the peaks are very narrow and weak.

* These are narrow as there are no crystal field effects on f-f transitions as the f-orbitals are

deeply buried inside the atom. The metal-ligand overlapping is very weak and hence no

vibronic coupling possible.

* These are also weak as there is no vibronic coupling. (The laporte forbidden transitions are

atleast possible due to vibronic coupling which destroy the symmetry of complex for a shortspan of time)

63.5) [Fe(bpy)3]2+ is expected to exhibit an MLCT rather than an LMCT absorption. Com-

ment.

Ans:- Iron is in the Fe(II) state which is harder to be reduced and hence LMCT is not observed.

MLCT is possible because metal d-orbitals are close in energy to the ligand * orbitals.

Remember this occurs in visible region.

63.6) Why is a transition from a t2gto an egorbital spin allowed in [V(OH2)6]3+?Ans:- The electronic configuration if V3+ion is t

2g

2eg

0. The transition is allowed as it involves NO

spin-flip. i.e., transition is of triplet to triplet type and hence allowed.

spin allowed

triplet triplet

t2g

eg

63.6) The reaction below produces both red and green crystals with

following data:

NiBr2+ PEtPh

2------> NiBr

2(PEtPh

2)

2( both red and green forms)

The following data is available:

Form data

Redeff

= 0

Greeneff

= 3.0 BM

The red form has zero dipole moment, whereas the green form has net dipole mo-

ment.When the red form is dissolved in dichloromethane, the magnetic moment of the

solution observed will be 2.69 BM.

Propose the possible geometries of both the forms.

Ans:- The four-coordinate compound can exist in two geometries:

i.e., square planar and tetrahedral

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As a square planar complex, the d8Ni(II) would be diamagnetic. Therefore the red com-

plex is square planar. It must be a trans isomer as its dipole moment is zero.

As a tetrahedral complex, the d8Ni(II) would have the electronic configuration (eg)4(t

2g)4

and have 2 unpaired electrons. Therefore the green form is Tetrahedral.

dxz dyz

dz

2

dxy

dx2

-y2

e

t2

Square planar Tetrahedral

A square planar complex will have a large crystal field splitting energy (delta) and absorb

shorter wavelengths and reflect long wavelengths such as red. A Tetrahedral complex will

have a small delta and absorb longer wavelengths and reflect shorter wavelengths such as

green.

As the tetrahedral form is more stable, the square planar complex is converted to tetrahe-

dral. The slightly low value of 2.69 BM may indicate there is an equilibrium between square

planar and tetrahedral forms.

63.7) When an ethanolic solution of NiBr2

is treated with two equivalents of the phosphine

PEtPh2, two products A and B can be obtained by varying the crystallisation condi-

tions.

Compound A has no dipole moment and is diamagnetic. One peak due to Ni-Br

stretching can be observed in its IR spectrum.

Compound B has a dipole moment and is paramagnetic. The IR spectrum of B

shows two peaks due to Ni-Br stretching.

By contrast, the analogous reaction with PtBr2and two equivalents of the same

phosphine PEtPh2gives just one compound, C, that has similar spectroscopic and

magnetic characteristics to A.Identify A, B and C and explain the observations.

Ans:- A is square planar and trans isomer. B is tetrahedral. (see the explanation given for the

previous question)

gkuryue854wycvli8794e5hggbli[097465eyv,mk[p9-7634utdglkok

][-\=--0=]90y8oedou760656w53w57328\

==-

\pob,jgd898

\=-pho8706w42yli

\=63.8) 1,10-Phenanthroline is added during the quantitative analysis of Fe(II) by UV-visible

spectroscopy. Explain why?

Ans:- The absorption bands of many transition metal ions in aqueous solutions are broad &

strongly influenced by the chemical environment, because of the spatial shape & orientation

of the d-orbitals. Such bands are rarely intense enough(low molar absorptivity) to use

directly for quantitative analysis. Hence the molar absorptivity is greatly augmented by

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Ans:- The mechanism and the equilibrium constant values for each step during the reaction are

given below.

2+ - +

2 6 (aq) (aq) 2 5 (aq) 2 (l) 1[Cd(OH ) ] + Br [CdBr(OH ) ] + H O ; logK 1.56

+ -

2 5 (aq) (aq) 2 2 4 (aq) 2 (l) 2[CdBr(OH ) ] + Br [CdBr (OH ) ] + H O ; logK 0.54

- -

2 2 4 (aq) (aq) 3 2 3 (aq) 2 (l) 3[CdBr (OH ) ] + Br [CdBr (OH ) ] + H O ; logK 0.06

- - 2-

3 2 3 (aq) (aq) 4 (aq) 2 (l) 4[CdBr (OH ) ] + Br [CdBr ] + 3H O ; logK 0.37

octahedral tetrahedral

It is clear from above data that, in the fourth step the octahedral aquo complex is con-

verted to tetrahedral all halo complex and 3 moles of water molecules are released (entropy

favored). Hence the K4is greater than K

3.

Note: Generally aquo complexes of M2+ions are six coordinated, whereas halo complexes

are four coordinated. (only generalization)

More explanation:Usually the stepwise formation constants lie in the order:

K1> K2> K3....... > Kn.This general trend can be explained simply by considering the decrease in the number of

the ligand H2O molecules available for replacement in susequent steps.

In general, Kn> K

n+1. But this order is reversed, when there is some structural change

during that step.

67) The number of faces and edges in IF7polyhedron are, respectively

1. 15 and 15

2. 10 and 15

3. 10 and 104. 15 and 10

Explanation:

F F

F

F F F

F

I

There are 10 faces and 15 edges.

69) O2can be converted to O

2

+by using

1. PtF6

2. KF

3. Na2S

2O

3

4. Br2

Explanation:* PtF

6is a strong oxidising agent and can remove electron from dioxygen.

O2+ PtF

6 ------>O

2

+[PtF6]-

* Bartlett made this compound, which lead to the discovery of first compound of noble gas,

Xenon. Bartlett realized that the first ionization energy of dioxygen, 1180 kJ mol-1is almost

equal to that of xenon, 1170 kJ mol-1. Furthermore, the dioxygen cation should be roughly the

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same size as a Xe+ion, and hence the lattice energies of the corresponding compounds should

be similar.

Xe + PtF6 ------>Xe[PtF

6]

71) CFSE of transition metal complexes can be determined by

1. UV-visible spectroscopy

2. IR spectroscopy

3. Microwave spectroscopy

4. NMR spectroscopy

Explanation:

* Crystal Field Stabilization Energy (CFSE) is calculated by using UV-visible spectroscopy.

First the max for a d-d electronic transition is recorded using spectrometer. The energy

corresponding to max is given bymax

hcE=

. This is equal to crystal field splitting energy ( ).

From this CFSE is calculated as follows for simple cases.

For octahedral complexes, CFSE = (-0.4a + 0.6b) x o

Where a = no. of electrons in t2g

orbitals

b = no. of electrons in egorbitals

For tetrahedral complexes, CFSE = (-0.6a + 0.4b) xt

Where a = no. of electrons in e orbitals

b = no. of electrons in t2orbitals

Note: max refers to the wavelength of peak with maximum molar absorptivity (

)

Additional questions:

71.1) Calculate the CFSE (crystal field stabilization energy) of [TiCl6]3-if the max is 770

nm.

Ans:- The energy corresponding to max = 770 nm can be calculated as

-34 8 -1

-9

max

-19

-22

-22 23

hc 6.625 x 10 J.sec x 3 x 10 m.secE= =

770 x 10 m

= 2.58 x 10 J

= 2.58 x 10 kJ per one molecule

= 2.58 x 10 kJ x 6.023 x 10 =155 kJ per mole

This energy is equal to the magnitude of energy se

paration( o ) between t2gand egin the

octahedral complex [TiCl6]3-.

E.C of Ti3+in the complex is 3d1and the only electron occupies the t2g

.

Hence the CFSE=1 x -2/5o

= -2/5 x 155 = 62 kJ. mole-1.

Practice questions

1) The UV-Vis spectrum of ReF6exhibits absorption at 32,500 cm-1. Calculate o in kJ mol

-1.

2) Which of the following is not a source of visible radiation?

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a) Deuterium lamp b) Tungsten lamp

c) Xenon arc lamp d) All.

Note: Deuterium and hydrogen lamps are essentially identical. But D2lamps are a little

brighter and hence are exclusively used over H2lamps.

73) In aqueous medium a mixture of KI and I2converts thiosulfate to

1. S4O62

2. SO42

3. S2O62

4. S2O42

Explanation:

* Iodine oxidises thiosulfate to tetrathionate.

2S2O

3

2- + I2

S4O

6

2- + 2I-

S O-

S

O

O- S S

O

O

O-

S S O-

O

O

thiosulfate tetrathionate

KI

KI reacts with I2by forming KI

3and increases its solubility in water.

* This reaction is used in iodometry.

Additional information:

Iodine is used in iodimetry and iodometry to estimate various oxidising and reducing

analytes.

Molecular Iodine(I2) is slightly soluble in water, but its solubility is greatly enhanced by

complexation with iodide. Hence when used as titrant, I2is dissolved in water containing

excess of KI. This solution has brown color.

I2 + I- ---------> I

3

-

To detect the end point in titrations using iodine, starch solution is added. Iodine imparts

navy blue color to starch solutions. The blue colored starch-iodine complex contains long

linear chains of I5-

, [I-I-I-I-I]-

in amylose part of starch.

IODIMETRY

* In iodimetry, reducing analyte is titrated directly with Iodine. T

he reducing analyte reduces

iodine(I2) to iodide(I-). This method is used in estimation of Ascorbic acid, glucose, phospho-

rus acid etc.,

* Ascorbic acid is oxidised by iodine to dehydroascorbic acid.

* Glucose is oxidised to gluconic acid (the -CHO group in glucose is converted to -COOH)

RCHO + 3OH-+ I3

- RCOO-+ 2H2O + 3I-

* Phosphorus acid is converted to phosphoric acid.

H3PO

3+ H

2O + I

3

- H3PO

4+ 2H++ 3I-

* In iodimetry, the starch solution can be added at the begining of the titration.

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IODOMETRY:

* In iodometry, first an oxidizing analyte is added to excess of I -solution to liberate I2which

is then titrated against a standard solution of thiosulfate.

first step ---- oxidation of I-to I2 by an oxidising analyte.

second step ---- reduction of I2to I-by thiosulfate.

Iodometry is used in the estimation of CuSO4, K

2Cr

2O

7, KMnO

4, Br

2, Cl

2, BrO

3

-etc.,

Estimation of copper:

CuSO4is made to react with excess of I-solution. Thus formed I

2is titrated with standard

thiosulfate solution.

2S2O

32- + I

3- S

4O

62- + 3I-

2CuSO4 + 4I- 2CuI + SO

42- + I

2white ppt

The iodine liberated in the first step is equivalent to CuSO4. The amount of CuSO4isestimated indirectly by estimating this liberated I

2with standard thiosulfate solution.

* In iodometry, starch solution is added only just before the equivalence point (which is

detected visually by fading of brown color of I3

-ions). This is because, at high concentrations,

some I2remains bound to starch particles.

* In the estimation of copper, the precipitated CuI tends to bind some iodine. To replace this

iodine and bring it into solution, thiocyanate solution is added at the end point.

Additional questions:

73.1) Copper(I) iodide is a stable species, while coppe(II) iodide does not exist. Explain.

Ans:- Cu2+can oxidise I-to I2and hence Copper(II) iodide does not exist.

2Cu2++ 2I- --------> 2Cu++ I2

Note:

1) In general Cu(II) ion is more stable than Cu(I) ion - (contrary to above observation).

2) Remember, other dihalides of copper (CuF2, CuCl

2and CuBr

2) are possible (Why?

Simply because other halides cannot be oxidized by Cu(II))

73.2) The black precipitate formed when hypo is added with silver n

itrate solution is

a) Ag2S b) Na

3[Ag(S

2O

3)

2] c) S d) Na

2S

Ans:- When less amount of hypo is added to silver nitrate, initially a white precipitate of silver

thiosulfate is formed and it is immediately reduced to black silver sulfide precipitate.

Na2S

2O

3+ 2AgNO

3Ag

2S

2O

3+ 2NaNO

3

Ag2S

2O

3+ H

2O Ag

2S + H

2SO

4

Black ppt.

Note: But when excess of hypo is used, the silver thiosulfate formed will be soluble in hypo

by forming a water soluble complex, sodium argento thiosulfate,Na3[Ag(S

2O

3)

2]

3Na2S

2O

3+ Ag

2S

2O

32Na

3[Ag(S

2O

3)

2]

H.W: What is the IUPAC name of above complex?

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1-n 1-n

M mole / litre=

sec sec

1-n 1-n n-1 -1= M .sec mole .L .sec

For 1storder reaction sec-1

For 2ndorder reaction L.mole-1sec-1

For 3rdorder reaction L2.mole-2sec-1

For zero order reaction mole .L-1sec-1

76) Heating is observed when N2gas at 200 atm is expanded at T > 600 K. It is because

1. inversion temperature is smaller than 600 K

2. N2is a real gas

3. Joule-Thomson coefficient is negative

4. Joule-Thomson coefficient is positive

Explanation:Joule-Thomson effect: The change in temperature of non-ideal gas when it is allowed to

expand adiabatically from high pressure to low pressure is called Joule-Thomson effect.

The temperature may increase or decrease during this process.

Joule-Thomson coefficient= lim 0JTH H

T Tas P

P P

The value of Joule-Thompson coefficient may be positive or negative. The temperature atwhich its sign is changed is called Joule-Thompson inversion temperature.

The sign of the Joule-Thompson coefficient is positive below the critical temperature and

the gas is cooled upon expansion. Hence below the critical temperature, the gas can be

liquified.

But the sign of the coefficient above the critical temperature is negative and the gas is

heated upon expansion. Hence above the critical tempertature, it is NOT possible to liquify

the gas.

Usually most of the gases are cooled upon adiabatic expansion at room temperature astheir critical temperatures are above the room temperature.

But the gases like H2, He and Ne have their critical temperatur

es below room temperature

(RT) and hence their temperature increases during adiabatic expansion at RT. These are also

difficult to liquify.

Answer to the question: inversion temperature is smaller than 600 K.

Practice questions:

1) The Joule-Thomson coefficient for N2is assumed to be a constant value of 0.15 K.bar-1. Ifthe N

2undergoes a drop of pressure of 200 bar, then the temperature will be:

1) raised by 30 K 2) dropped by 30 K

3) remains constant 4) no change in temperature.