Thermodynamics Test

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Thermodynamics Test - 8

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Question 1
1 / -0

Standard enthalpies of formation of CO₂ (g), CO (g), N₂O (g) and NO₂(g) in kJ/mol are -393, -110, 81 and 34, respectively. What is the value of ∆H (in kJ) in the following reaction?

2NO₂ (g) + 3CO (g) → N₂O (g) + 3CO₂ (g)

A
-836

B
-1460

C
1460

D
836

Solution

By using the formula:
D_H for the reaction = [Sum of heat of formation for products] - [Sum of heat of formation for reactants]

D_H for the reaction = [3(-393) + 81] - [3(-110) + 2(34)] = -836 kJ
Question 2
1 / -0

In the following question, there are two statements, Assertion (A) and Reason (R). Consider both the statements independently and mark your answer.

Assertion (A): There is a natural asymmetry between converting work to heat and converting heat to work.
Reason (R): It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equal amount of work by the system on the surroundings.

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

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

C
A is true, but R is false.

D
A is false, but R is true.

Solution

Both statements are true, and R is the correct explanation of A.
There is a natural asymmetry between converting work to heat and converting heat to work. The statement is based on the first and second laws of thermodynamics and implies that work can be completely converted into heat but the heat energy cannot be completely converted into work; hence the asymmetry.

Kelvin-Planck formulation of the second law of thermodynamics: It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equal amount of work by the system on the surroundings. The second statement supports the first statement.
Question 3
1 / -0

Gibbs combined the first and second laws of thermodynamics and gave a thermodynamic function 'G', which gets affected by

(a) the change in the enthalpy of the system
(b) the change in the entropy of the system
(c) the temperature of the system

A
Only (a)

B
Only (b)

C
All (a), (b) and (c)

D
Only (c)

Solution

'G' is the Gibb's energy of the system, i.e., the amount of energy that is available to the system to do useful work. The change in free energy during a process (∆ G) can be calculated by the Gibb's Helmholtz equation:

∆G = ∆H - T∆S
The process is spontaneous if the value of ∆ G is negative.
Question 4
1 / -0

Identify the correct statement for change of Gibbs energy for a system (∆G_system) at a constant temperature and pressure.

A
If ∆G_system > 0, then the process is spontaneous.

B
If ∆G_system= 0, then the system has attained equilibrium.

C
If ∆G_system= 0, then the system is still moving in a particular direction.

D
If ∆G_system< 0, then the process is not spontaneous.

Solution

The correct interpretation, from the value of ∆ G are:
If ∆G system < 0, then the process is spontaneous.
If ∆G system > 0, then the process is non-spontaneous.
If ∆G system = 0, then the process is in equilibrium.

Question 5

1 / -0

Match the columns:

Column – I	Column – II
(A) Adiabatic process	(a) –w = q
(B) Endothermic reaction	(b) minimum destruction of work
(C) Isothermal process	(c) q = 0
(D) Reversible compression	(d) ∆H > 0
	(e) ∆E = w
	(f) ∆E = 0

(A) – (c) and (e); (B) – (d); (C) – (a) and (f); (D) – (b)

(A) – (e) and (f); (B) – (d); (C) – (c) and (f); (D) – (a)

(A) – (e) and (f); (B) – (d); (C) – (a) and (c); (D) – (b)

(A) – (a) and (f); (B) – (d); (C) – (b) and (e); (D) – (c)

Solution

(A) Adiabatic process does not allow exchange of heat with the surroundings, i.e. q = 0.
Also, since work is done at the expense of internal energy. ∆E = w
So, (A) matches with (c) and (e).

(B) Endothermic reactions are accompanied by increase in enthalpy of the system, ∆H > 0. So, (B) matches with (d).

(C) For an isothermal process there is no change in the internal energy of the system, ∆E = q + w = 0. So, –w = q.
So, (C) matches with (a) and (f).

(D) Reversible compression is accompanied by destruction of work, but it is always less than that in an irreversible process. So, destruction of work is minimum in a reversible process.
So, (D) matches with (b).

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Thermodynamics Test - 8

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Thermodynamics Test - 8

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Question 1

-836

-1460

1460

836

Solution

Question 2

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

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

A is true, but R is false.

A is false, but R is true.

Solution

Question 3

Only (a)

Only (b)

All (a), (b) and (c)

Only (c)

Solution

Question 4

If ∆Gsystem > 0, then the process is spontaneous.

If ∆Gsystem = 0, then the system has attained equilibrium.

If ∆Gsystem = 0, then the system is still moving in a particular direction.

If ∆Gsystem < 0, then the process is not spontaneous.

Solution

Question 5

(A) – (c) and (e); (B) – (d); (C) – (a) and (f); (D) – (b)

(A) – (e) and (f); (B) – (d); (C) – (c) and (f); (D) – (a)

(A) – (e) and (f); (B) – (d); (C) – (a) and (c); (D) – (b)

(A) – (a) and (f); (B) – (d); (C) – (b) and (e); (D) – (c)

Solution

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If ∆G_system > 0, then the process is spontaneous.

If ∆G_system= 0, then the system has attained equilibrium.

If ∆G_system= 0, then the system is still moving in a particular direction.

If ∆G_system< 0, then the process is not spontaneous.