Entropy Chang...

Entropy Change in Irreversible Process

Trending Questions

Q. Following table represents critical temperature of some gases. Arrange these gases in their increasing order of liquefaction.

Gas	$H_{2}$	$H e$	$N_{2}$	$O_{2}$
$T_{c}$ $(K)$	33.2	5.3	126	154.3

$H e < N_{2} < H_{2} < O_{2}$
$H_{2} < H e < N_{2} < O_{2}$
$H e < H_{2} < N_{2} < O_{2}$
$O_{2} < N_{2} < H_{2} < H e$

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Q. For irreversible expansion of an ideal gas under isothermal condition, the correct option is :

$Δ \cup \neq 0, Δ S_{total} = 0$
$Δ \cup = 0, Δ S_{total} = 0$
$Δ \cup \neq 0, Δ S_{total} \neq 0$
$Δ \cup = 0, Δ S_{total} \neq 0$

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Q. Why is spontaneous process irreversible

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Q. The incorrect option for free expansion of ideal gas under isothermal condition is

$Δ H = 0$
$Δ S = 0$
$Δ T = 0$
$w = 0$

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Q. The total entropy of a system and its surrounding increases, if the process is

reversible
irreversible
exothermic
endothermic

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Q. Assertion:
Critical temperature of

C O_{2}

304 K

, it cannot be liquified above

304 K

.
Reason:
At a certain temperature,

Volume \propto \frac{1}{Pressure}

Both assertion and reason are true and reason is the correct explanation of assertion.
Both assertion and reason are true but reason is not the correct explanation of assertion.
Assertion is true but reason is false.
Both assertion and reason are false.

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1 mole of a triatomic non-linear ideal gas is subjected to an adiabatic process from initial temperature 400 K and pressure 32 atm to final pressure 2 atm. If the expansion is carried out against vaccum, the entropy change (in cal $K^{- 1}$ ) for the system is $[l n 2 = 0.7]$

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Q. What is the cell entropy change of the following cell ?

P t (s) | H_{2} (g, 1 a t m) | C H_{3} C O O H, H C l | | K C l (a q) | H g_{2} C l_{2} (s) | H g

EMF of the cell is found to be

0.0045 V

at 298 K and temperature coefficient is

3.4 \times 10^{- 4} V K^{- 1}

$105 k J K^{- 1}$
$205.7 J K^{- 1}$
$5.45 J K^{- 1}$
$65.61 J K^{- 1}$

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Q. Two moles of an ideal gas are expanded irreversibly and isothermally at

37^{\circ} C

until the volume is doubled, and 3.41 kJ heat is absorbed from the surroundings.

Δ S_{t o t a l} (system + surrounding)

(R = \frac{25}{3} J/molK, ln 2=0.7, ln 3 = 1.1)

- 0.667 J/K
0.667 J/K
22.7 J/K
\N

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Q. Given that:

S_{H_{2}}^{\circ} = 131 J K^{- 1} m o l^{- 1} S_{C l_{2}}^{\circ} = 223 J K^{- 1} m o l^{- 1} a n d S_{H C l}^{\circ} = 187 J K^{- 1} m o l^{- 1}

The standard entropy change in formation of

1

mole of

H C l (g)

from

H_{2} (g)

and

C l_{2} (g)

will be:
(Given, reaction for formation of

H C l

H_{2} + C l_{2} \to 2 H C l

)

$187 J K^{- 1}$
$374 J K^{- 1}$
$20 J K^{- 1}$
$10 J K^{- 1}$

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Q. Can anyone say what is reversible inhibition and irreversible inhibition?

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Q. Two moles of an ideal gas are expanded irreversibly and isothermally at

37^{\circ} C

until the volume is doubled, and 3.41 kJ heat is absorbed from the surroundings.

Δ S_{t o t a l} (system + surrounding)

(R = \frac{25}{3} J/molK, ln 2=0.7, ln 3 = 1.1)

- 0.667 J/K
0.667 J/K
22.7 J/K
\N

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Q. Starting with the same initial conditions, one mole of an ideal monoatomic gas expands reversibly from volume

V_{1}

V_{2}

in two different ways. The work done by gas is

w_{1}

, if the process is purely isothermal,

w_{2}

if purely isobaric and

w_{3}

if purely adiabatic, then:

$w_{2} > w_{1} > w_{3}$
$w_{2} > w_{3} > w_{1}$
$w_{1} > w_{2} > w_{3}$
$w_{1} > w_{3} > w_{2}$

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Q. The correct folrmula of Q by using Born-Haber cycle for the preparation of

M g B r_{2} (s)

will be:
S

=

heat of sublimation of Mg(s)
I

=

ionisation energy of Mg(g)
D

=

dissociation energy of Br

_{2}

(g)
-E

=

electron affinity of Br(g)
-U

=

lattice energy of MgBr

_{2}

(s)
Q

=

heat of formation of MgBr

_{2}

(s)

$Q = S + I + D - 2 E - U$
$Q = S - I + D - 2 E - U$
$Q = S + I + D - E - 2 U$
None of these

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A diatomic ideal gas initially at 272 K is given 100 cal heat due to which system did 209 J work. Molar heat capacity ( $C_{m}$ ) of gas for the process is

5/2 R
5R
5/4 R
3/2 R

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Q. For a chemical reaction

A + B ⇌ C + D (Δ_{r} H^{o} = 80 k J m o l^{1})

the entropy change

Δ_{r} S^{o}

depends on the temperature

T

(in

K

) as

Δ_{r} S^{o} = 2 T (J K^{- 1} m o l^{- 1})

. Minimum temperature at which it will become spontaneous is ________

K

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Two moles of ideal monoatomic gas was taken through isochoric heating from 100 K to 800 K. If the process is carried out irreversibly (one step) then $Δ S_{s y s t e m} + Δ S_{s u r r o u n d i n g}$ is

[Given : $l n 2 = 0.7$ and $R = 2 c a l / m o l K$ ]

0
3.675 cal/K
7.35 cal/K
None of these

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Q. For a spontaneous process, which of the following is true?

$Δ S_{s y s}$ is positive
$Δ S_{s u r r}$ is positive
$Δ S_{t o t a l}$ is positive
$Δ S_{t o t a l}$ is negative

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Q. When

1

mol of an ideal gas is compressed to half of its volume, its temperature becomes double; then the change in entropy (

Δ S

) would be:

$C_{V} ln 2$
$C_{P} ln 2$
$(C_{V} - R) ln 2$
$C_{V} R ln 2$

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Identify the correct statement regarding a spontaneous process.

Lowering of energy in the process is the only criterion for spontaneity.
For a spontaneous process in an isolated system, the change in entropy is positive.
Exothermic process are always spontaneous.
Endothermic process are never spontaneous.

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Q. Two mole of an ideal gas is expanded irreversibly and isothermally at 37

^{\circ}

C until its volume is doubled and 3.41 kJ heat is absorbed from surrounding.

Δ S_{t o t a l}

(system + surrounding) is :

$0.52 J / K$
$0$
$22.52 J / K$
none of the above

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Q. Two mole of an ideal gas is expanded irreversibly and isothermally at

37^{\circ} C

until its volume is doubled and

3.41

kJ heat is absorbed from surrounding.

Δ S_{t o t a l}

(system + surrounding) is :

$- 0.52$ J/K
$0.52$ J/K
$22.52$ J/K
$0$

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1

mole of an ideal gas is allowed to expand isothermally at

27^{o} C

until its volume is tripled. Calculate

Δ S_{s y s}

and

Δ S_{u n i v}

under the following conditions:
(a) the expansion is carried out reversibly.
(b) the expansion is a free expansion

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Pressure of 10 moles of an ideal gas is changed from 2 atm to 1 atm against constant external pressure without change in temperature. If surrounding temperature (300 K) and pressure (1 atm) always remains constant then calculate total entropy change ( $Δ S_{s y s t e m} + Δ S_{s u r r o u n d i n g})$ for given process. [Given: ℓn2 = 0.70 amd R=8.0 J/mol/K]

56 J/K
14 J/K
16 J/K
None of these

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Q. Which of the following statements is correct?

${(\frac{\partial H}{\partial T})}_{p} - {(\frac{\partial U}{\partial T})}_{v} = R$
${(\frac{\partial H}{\partial T})}_{p} > {(\frac{\partial U}{\partial T})}_{v}$
${(\frac{\partial U}{\partial V})}_{T}$ for ideal gas is zero
All of the above

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For a spontaneous process.

$Δ G_{s y t e m}$ = +ve only
$Δ G_{s y t e m}$ = zero
$Δ S_{t o t a l}$ = - ve
$Δ S_{t o t a l}$ = +ve

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Q. Select correct option regarding change in entropy of system

(Δ S_{s y s})

and of the surroundings

(Δ S_{s u r r})

$Δ S_{s y s} > 0, Δ S_{s u r r} < 0$
$Δ S_{s y s} < 0, Δ S_{s u r r} > 0$
$Δ S_{s y s} = 0, Δ S_{s u r r} = 0$
$Δ S_{s y s} > 0, Δ S_{s u r r} = 0$

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Q. Entropy change of a system is determined by the initial and final states only, irrespective of how the system has changed its states.
If true enter 1 else 0

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Q. Two moles of an ideal gas are expanded irreversibly and isothermally at

37^{\circ} C

until the volume is doubled, and 3.41 kJ heat is absorbed from the surroundings.

Δ S_{t o t a l} (system + surrounding)

(R = \frac{25}{3} J/molK, ln 2=0.7, ln 3 = 1.1)

- 0.667 J/K
0.667 J/K
22.7 J/K
0

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[Given : $l n 2 = 0.7$ and $R = 2 c a l / m o l K$ ]

0
3.675 cal/K
7.35 cal/K
None of these

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