Thermodynamics Test

Result

Thermodynamics Test - 15

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Born - Haber cycle may be used

A
to find out electron affinity of non-metal atoms

B
to find out lattice energy of the ionic compounds

C
for the preparation of ammonia in industries

D
both A and B

Solution

Born - Haber cycle is used to calculate the lattice energy for one mole of the ionic compound formed. According to this cycle, the value of lattice energy and enthalpy of formation is equal to the summation of sublimation energy of the metal atom, Ionization energy required for the formation of the cation, enthalpy of atomization and electron affinity.
From this formula, we can calculate any of the above values.
Question 2
1 / -0

Gibbs energy change $\Delta$ G is related to equilibrium constant K as:

A
$\Delta G^{0}$ = $- RT lnk$

B
$\Delta G^{0}$ = $RT lnk$

C
$lnk$ = $-\frac{RT}{\Delta G^{0}}$

D
$lnk$ = $\frac{\Delta G^{0}}{RT}$

Solution

$\Delta G^{o}=-RT$ Ink

Gibbs energy change $\Delta G$ is related to equilibrium constant K as above
Question 3
1 / -0

Which of the following relation is incorrect?

A
$\Delta S=\Delta H$ + $T\Delta G$

B
$\Delta S=\frac{\Delta H-\Delta G}{T }$

C
$– \Delta G = - W_{non\ exp.}$

D
$W_{PV} = –P \Delta V$

Solution

$\Delta G=\Delta H-T\Delta S$

$\therefore \Delta S=\frac{\Delta H-\Delta G}{T}$

$\Delta G=-W_{non\ exp.}$

Gibbs free energy is equal to useful work for a spontaneous process

$P-V$ work $=-P\Delta V$
Question 4
1 / -0

Assertion (A): The value of $\Delta E$ is negative if the internal energy of a system increases.
Reason (R): The value of $Q$ is negative if heat is absorbed by the system.

A
Both A and R are true and R is the correct explanation of A.

B
Both A and R are true but R is not the correct explanation of A.

C
A is true but R is false.

D
Both A and R are false

Solution

The value of $\Delta E$ is positive if the internal energy of a system increases. The value of $\Delta Q$ is positive, if heat is absorbed by the system.
Question 5
1 / -0

A heat engine absorbs heat $Q_{1}$ at temperature $T_{1}$ and heat $Q_{2}$ at temperature $T_2$ . Work done by the engine is $(Q_{1}+Q_{2}) J$ . This data:

A
violates $1^{st}$ law of thermodynamics

B
violates $1^{st}$ law of thermodynamics $Q_{1}$ is $-$ ve

C
violates $1^{st}$ law of thermodynamics $Q_{2}$ is $-$ ve

D
does not violates $1^{st}$ law of thermodynamics

Solution

According, First Law of Thermodynamics,

Change in $E = Q +$ work done

Change in internal energy is zero for heat engines because initial and final states are same.

Total work done $= -$ Total heat exchange $= -(Q_1+Q_2)$ .
Question 6
1 / -0

Correct relation among the following.

A
$\Delta G_{system} = \Delta S_{total}$

B
$\Delta G_{system}=\Delta H_{system}-T\Delta S_{system}$

C
$\Delta$ G = $\Delta$ H + T $\Delta$ S

D
$\Delta S=\frac{\Delta G+\Delta H}{\Delta T}$

Solution

$\Delta G_{system}=T\Delta S_{total}$
$\Delta G_{system}=\Delta H_{system}-T\Delta S_{system}$
If an isolated system (Q $=$ 0) is at constant pressure (Q $=\Delta H$ ) then
$\Delta H=0$
Therefore, the Gibbs free energy of an isolated system is $\Delta G_{system}=-T\Delta S_{total}$
Question 7
1 / -0

Gibbs -Helmholtz equation is______

A
$\Delta$ G= $\Delta$ H-T $\Delta$ S

B
$\Delta$ G= $\Delta$ H- $\Delta$ S

C
$\Delta$ G =T $\Delta$ H-T $\Delta$ S

D
$\Delta$ G=T $\Delta$ H- $\Delta$ S

Solution

$\Delta G=\Delta H-T\Delta S$

is the Gibbs Helmholtz equation.

Option A is correct.
Question 8
1 / -0

Which of the following statements is incorrect?

A
It is not possible to compress a gas at a temperature below $T_c$

B
At a temperature below $T_c$ , the molecules are close enough for the attractive forces to act, and condensation occur

C
No condensation takes place above $T_c$

D
The kinetic energy of the gas molecules is higher above $T_c$ , and the attraction between them decreases

Solution

Critical temperature of the substance is the temperature at a substance's critical point.
Above $T_c$ we cannot liquify gas keep pressure constant although on $\uparrow$ pressure we can liquify.

Question 9

1 / -0

Identify the correct statement for change of Gibbs energy for a system (

\Delta

_{system}

) at constant temperature and pressure:

If $\Delta$ G $_{system}$ > 0, the process is spontaneous

If $\Delta$ G $_{system}$ = 0, the system has attained equilibrium

If $\Delta$ G $_{system}$ = 0, the system is still moving in a particular direction

If $\Delta$ G $_{system}$ < 0, the process is not spontaneous

Solution

At equilibrium, the entropy of system is maximum. Therefore change in entropy i.e;

\triangle S=0

Process	Sign of $\triangle S$	Sign of $\triangle G$
Spontaneous	Positive, $\triangle S_{total} >0$	Negative, $\triangle G <0$
Non-Spontaneous	Negative, $\triangle S_{total} < 0$	Positive, $\triangle G>0$
Equilibrium	$\triangle S_{total}=0$	$\triangle G=0$

Question 10
1 / -0

Cell reaction is spontaneous when :

A
G is negative

B
G is positive

C
E is positive

D
E is negative

Solution

Hint: The Gibbs free energy is related to the spontaneity of a reaction.

Explanation: If the Gibbs free energy is negative then the reaction is spontaneous.

$\Delta {G^ \circ } = - nF{E^ \circ }$

For $\Delta {G^ \circ } < 0$ , the reaction, ${E^ \circ }_{cell}$ should be positive so the correct answer is (A)

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 15

Result

Thermodynamics Test - 15

Score

-

Rank

-

SHARING IS CARING

Question 1

to find out electron affinity of non-metal atoms

to find out lattice energy of the ionic compounds

for the preparation of ammonia in industries

both A and B

Solution

Question 2

ΔG0\Delta G^{0}ΔG0 = −RTlnk- RT lnk−RTlnk

ΔG0\Delta G^{0}ΔG0 = RTlnkRT lnkRTlnk

lnklnklnk = −RTΔG0-\frac{RT}{\Delta G^{0}}−ΔG0RT​

lnklnklnk = ΔG0RT\frac{\Delta G^{0}}{RT}RTΔG0​

Solution

Question 3

ΔS=ΔH\Delta S=\Delta HΔS=ΔH + TΔGT\Delta GTΔG

ΔS=ΔH−ΔGT\Delta S=\frac{\Delta H-\Delta G}{T }ΔS=TΔH−ΔG​

–ΔG=−Wnon exp.– \Delta G = - W_{non\ exp.}–ΔG=−Wnon exp.​

WPV=–PΔ VW_{PV} = –P \Delta VWPV​=–PΔ V

Solution

Question 4

Both A and R are true and R is the correct explanation of A.

Both A and R are true but R is not the correct explanation of A.

A is true but R is false.

Both A and R are false

Solution

Question 5

violates 1st1^{st}1st law of thermodynamics

violates 1st1^{st}1st law of thermodynamics Q1Q_{1}Q1​ is −-−ve

violates 1st1^{st}1st law of thermodynamics Q2Q_{2}Q2​ is −-−ve

does not violates 1st1^{st}1st law of thermodynamics

Solution

Question 6

ΔGsystem=ΔStotal\Delta G_{system} = \Delta S_{total}ΔGsystem​=ΔStotal​

ΔGsystem=ΔHsystem−TΔSsystem\Delta G_{system}=\Delta H_{system}-T\Delta S_{system}ΔGsystem​=ΔHsystem​−TΔSsystem​

Δ\Delta ΔG =Δ\Delta Δ H + T Δ\Delta ΔS

ΔS=ΔG+ΔHΔT\Delta S=\frac{\Delta G+\Delta H}{\Delta T}ΔS=ΔTΔG+ΔH​

Solution

Question 7

Δ\Delta ΔG=Δ\Delta Δ H-T Δ\Delta Δ S

Δ\Delta ΔG=Δ\Delta ΔH- Δ\Delta ΔS

Δ\Delta ΔG =TΔ\Delta Δ H-TΔ\Delta Δ S

Δ\Delta ΔG=T Δ\Delta Δ H- Δ\Delta Δ S

Solution

Question 8

It is not possible to compress a gas at a temperature below TcT_cTc​

At a temperature below TcT_cTc​, the molecules are close enough for the attractive forces to act, and condensation occur

No condensation takes place above TcT_cTc​

The kinetic energy of the gas molecules is higher above TcT_cTc​, and the attraction between them decreases

Solution

Question 9

If Δ\DeltaΔGsystem_{system}system​ > 0, the process is spontaneous

If Δ\DeltaΔGsystem_{system}system​ = 0, the system has attained equilibrium

If Δ\DeltaΔGsystem_{system}system​ = 0, the system is still moving in a particular direction

If Δ\DeltaΔGsystem_{system}system​ < 0, the process is not spontaneous

Solution

Question 10

G is negative

G is positive

E is positive

E is negative

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$\Delta G^{0}$ = $- RT lnk$

$\Delta G^{0}$ = $RT lnk$

$lnk$ = $-\frac{RT}{\Delta G^{0}}$

$lnk$ = $\frac{\Delta G^{0}}{RT}$

$\Delta S=\Delta H$ + $T\Delta G$

$\Delta S=\frac{\Delta H-\Delta G}{T }$

$– \Delta G = - W_{non\ exp.}$

$W_{PV} = –P \Delta V$

violates $1^{st}$ law of thermodynamics

violates $1^{st}$ law of thermodynamics $Q_{1}$ is $-$ ve

violates $1^{st}$ law of thermodynamics $Q_{2}$ is $-$ ve

does not violates $1^{st}$ law of thermodynamics

$\Delta G_{system} = \Delta S_{total}$

$\Delta G_{system}=\Delta H_{system}-T\Delta S_{system}$

$\Delta$ G = $\Delta$ H + T $\Delta$ S

$\Delta S=\frac{\Delta G+\Delta H}{\Delta T}$

$\Delta$ G= $\Delta$ H-T $\Delta$ S

$\Delta$ G= $\Delta$ H- $\Delta$ S

$\Delta$ G =T $\Delta$ H-T $\Delta$ S

$\Delta$ G=T $\Delta$ H- $\Delta$ S

It is not possible to compress a gas at a temperature below $T_c$

At a temperature below $T_c$ , the molecules are close enough for the attractive forces to act, and condensation occur

No condensation takes place above $T_c$

The kinetic energy of the gas molecules is higher above $T_c$ , and the attraction between them decreases

If $\Delta$ G $_{system}$ > 0, the process is spontaneous

If $\Delta$ G $_{system}$ = 0, the system has attained equilibrium

If $\Delta$ G $_{system}$ = 0, the system is still moving in a particular direction

If $\Delta$ G $_{system}$ < 0, the process is not spontaneous