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tatyana61 [14]
3 years ago
13

The 1996 Nobel Prize in Physics was awarded to Douglas Osheroff, Robert Richardson, and David Lee for discovering superfluidity

(apparently frictionless flow) in 3 He. When 3 He is cooled to 2.7 mK, the liquid settles into an ordered superfluid state. Predict the sign of the entropy change for the conversion of liquid 3 He into its superfluid state.
Chemistry
1 answer:
aleksklad [387]3 years ago
4 0

Explanation:

As it is known that entropy is the degree of randomness of the molecules of a substance or object.

When particles or molecules of a substance are placed in orderly manner then it means that there is low entropy. But when molecules are randomly moving then it means the system is in disorder and will have high entropy.

Since, molecules of a solid are orderly arranged. Therefore, solid substances have very low entropy.

Liquid molecules have the ability to slide past each other. Hence, liquids have moderate entropy whereas gas molecules are held by weak intermolecular force.

So, molecules of a gas move much more rapidly and hence, they have high entropy.

As it is given that 3 He is cooled to 2.7 mK, and the liquid settles into an ordered super-fluid state.

Therefore, there will occur a decrease in entropy.

Thus, we can conclude that the sign of entropy for the given case is negative.

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Answer:

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Explanation:

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Why is the nitrogen group considered a mixed group?
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From the following reaction and data, find (a) S o of SOCl2 (b) T at which the reaction becomes nonspontaneous SO3(g) + SCl2(l)
disa [49]

Answer:

618 J/Kmol

T > 1.36 x 10³ K

Explanation:

The  balanced reaction of interest is:

                           SO₃ (g) + SCl₂ (l) ⇒    SOCl₂ (l) +     SO₂ (g)

with the data:

ΔHºf (kJ/mol)      -396          -50.0          -245.6         -296.8

Sº(J/mol·K)             256.7       184               ?               -248.1

ΔGº=  -75.2 kJ

We know, we can find the standard  change inGibb´s free energy from the equation:

ΔGºrxn =  ΔHºrxn - TΔSºrxn

So we can calculate ΔHºrxn  = ∑ ΔHºf prod  -  ΔHºreact, and substitute into this equation to solve Sº SOCl₂.

ΔHºrxn = ( -245.6 + (-296.8) ) - ( -396 - 50) kJ = - 96.4 kJ

Similarly  for ΔSºrxn

 ΔSºrxn = (-0.248.1 +Sº SOCl₂) - (0.256.7 +0.184) kJ/K

= -0.689 kJ /K -+ Sº SOCl₂

Plugging the values for the expression for  ΔGºrxn:

-75.2 kJ = -96.4 kJ - 298 K  x  ( -0.689 kJ /K + Sº SCl₂ )

-75.2 kJ = -96.4 kJ + 205.3 Kj - 298 Sº SCl₂

-184  kJ = -298 K  x Sº SCl₂

0.618 kJ/molK = Sº SCl₂

= 0.618 kJ/K x 1000 J = 618 J/Kmol

For the second part we will still be using the Gibb´s free energy change  equation as above , but now we will solve for T when the reaction becomes  non-spontaneous.

For the reaction to become non-spontaneous  ΔGº is positive, and this happens when the term  TΔSº becomes greater tha ΔHº:

ΔGºrxn =  ΔHºrxn - TΔSºrxn

0 =   ΔHºrxn - TΔSºrxn ⇒  TΔSºrxn  =  ΔHºrxn

                                           T= ΔHºrxn / ΔSºrxn

ΔSºrxn  = -0.689 J/Kmol + 0.618 J/Kmol = -0.0710 kJ/Kmol

( using the value  the value just calculated from above )

T =  - 96.4 kJ / -0.071 kJ/K = 1.36 x 10³ K

For temperatures greater than 1.36 x 10³ K the reaction becomes non-spontaneous.

4 0
4 years ago
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