Thermodynamics Test

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Thermodynamics Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

The statement "The change of enthalpy of a chemical reaction is same whether the reaction takes place in one or several steps" is:

A
Le Chatelier's law

B
van't Hoff's law

C
first law of thermodynamics

D
Hess's law.

Solution

Hess's law of constant heat summation states that, "The standard enthalpy of a reaction, which takes place in several steps, is the sum of the standard enthalpies of intermediate reactions into which the overall reactions may be divided at the same temperature".
Question 2
1 / -0

1 mole of an ideal gas STP is subjected to a reversible adiabatic expansion to double its volume. The change in internal energy $$(\gamma = 1.4)$$

A
1169 J

B
769 J

C
1373 J

D
969 J

Solution

Given,
$$\gamma=1.4$$
$$n=1$$
At STP, $$T_1=273.15K$$
$$V_1=V$$
$$V_2=2V$$
In adiabatic process,
$$PV^{\gamma}=constant=k$$. . . . . . . .(1)
From ideal gas equation,
$$PV=nRT$$
$$P=\dfrac{1\times RT}{V}=\dfrac{RT}{V}$$. . . . .. (2)
Substitute equation (2) in equation (1), we get
$$\dfrac{RT}{V}V^{\gamma}=k$$
$$TV^{\gamma-1}=k'(constant)$$
$$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$$
$$T_2=T_1(\dfrac{V_1}{V_2})^{\gamma-1}$$
$$T_2=273.15\times (\dfrac{V}{2V})^{1.4-1}$$
$$T_2=207.00K$$
Change in internal energy for adibatic process,
$$\Delta U=\dfrac{R}{\gamma-1}(T_1-T_2)$$
$$\Delta U=\dfrac{8.31}{1.4-1}(273.15-207.00)$$
$$\Delta U=1373J$$
The correct option is C.
Question 3
1 / -0

For the reaction at $$\ { 25 }^{ o }C,{ X }_{ 2 }{ O }_{ 4(l) }\longrightarrow 2X{ O }_{ 2(g) }\quad $$
$$\Delta H=2.1kcal$$ and $$\Delta S=20cal\quad { K }^{ -1 }$$.
The reaction would be:

A
spontaneous

B
non-spontaneous

C
at equilibrium

D
unpredictable

Solution

Answer:- (A) Spontaneous
Given:-
T $$= 25^oC = (273 + 25)K = 298K$$
$$\Delta H = 2.1 kcal = 2.1 \times {10}^{3} cal \; \& \; \Delta S = 20cal{K}^{-1}$$
Gibbs free energy $$(\Delta G) = \Delta H - T \Delta S$$
$$\Delta G = 2.1 \times {10}^{3} - (298 \times 20) = 2100 - 5960 = -2860cal$$
$$\because \Delta G = negative$$, thus the reaction is spontaneous.
Question 4
1 / -0

Internal energy of an ideal gas depends upon

A
temperature only

B
volume only

C
both volume and temperature

D
neither volume nor temperature

Solution

Hint: $$U=nC_v\delta T$$
Explanation: The internal energy of a gas is an inherent property. For an ideal gas, there is the absence of inter-molecular collision. So, gas has only translational kinetic energy. So, internal energy doesn’t depend on external factors like volume and pressure. So, the internal energy of an ideal gas depends on temperature only.

$$Answer:$$

Hence, option A is the correct option.
Question 5
1 / -0

Three stars A, B, C have surface temperatures $$T_A$$, $$T_B$$ and $$T_C$$. A appears bluish, B appears reddish and C appears yellowish. We can conclude that:

A
$$T_A > T_C > T_B$$

B
$$T_A > T_B > T_C$$

C
$$T_B > T_C > T_A$$

D
$$T_C > T_B > T_A$$

Solution

According to Wien's Displacement law we know that:-
$$\lambda=\dfrac bT \ldots \ldots (1)$$.

Given that three stars $$A, B $$ and $$C$$ where $$A$$ appears bluish, $$B$$ appears reddish and $$C$$ appears yellowish.

We know that $$\lambda_{blue}< \lambda_{yellow}< \lambda_{Red}$$
$$\implies \lambda_{A}<\lambda_{C}<\lambda_{B} ..... (2)$$

from (1) and (2) we get
$$\implies T_{A} > T_{C}> T_{B}$$

Hence option $$(A)$$ is correct
Question 6
1 / -0

1 kg of water is heated from 40 C to 70 C, If its volume remains constant, then the change in internal energy is (specific heat of water = 4148 J$${ kg }^{ -1 }{ K }^{ -1 })$$

A
$$2.44\times { 10 }^{ 5 }$$J

B
$$1.62\times { 10 }^{ 5 }$$J

C
$$1.24\times { 10 }^{ 5 }$$J

D
$$2.62\times { 10 }^{ 5 }$$J

Solution

Since volume does not change as the energy supplied is used to change the internal energy.
$$\Delta E=\Delta U=mC\theta\\ \quad=1\times4148\times30\\ \quad=1.24\times 10^5 J$$
Question 7
1 / -0

An electric heater supplies heat to a system at a rate of 120 W. If system performs work at a rate of 80 J $${ s }^{ -1 }$$, the rate of increase in internal energy is :

A
30 J $${ s }^{ -1 }$$

B
40 J $${ s }^{ -1 }$$

C
50 J $${ s }^{ -1 }$$

D
60 J $${ s }^{ -1 }$$

Solution

Q=120 W
W=80 Js$$^{-1}$$
From the first law of thermodynamics:
$$Q=U+W$$
where U=internal energy
$$\therefore U=Q-W =120-80W=40 J s^{-1}$$
Question 8
1 / -0

An ideal gas undergoing a change of state from A to B through four different paths I, II ,III and IV as shown in the P-V diagram that lead to the same change of state, then the change in internal energy is

A
Same in I and II but not in III and IV

B
Same in III and IV but not in I and II

C
Same in I,II and III but not in IV

D
Same in all of the four cases

Solution

Internal energy is a function of the state of a system and does not depend on the path. Therefore, internal energy will be same.
Question 9
1 / -0

For the combustion of $$CH_4$$ at 1 atm pressure & 300 K, which of the following options is correct?
$$2M + O_2 \xrightarrow 2MO$$
$$\Delta G_{1000 ^{\circ}C} = -921 kJ/mol$$
$$\Delta G_{1900 ^{\circ}C} = -300 kJ/mol$$
$$2C + O_2 \xrightarrow 2CO$$
$$\Delta G _{1000 ^{\circ}C} = -432 kJ/mol$$
$$\Delta G_{1900 ^{\circ}C} = -624 kJ/mol$$
$$MO + C \xrightarrow{\Delta} M + CO \uparrow$$
This reaction is feasible at temperature:

A
$$1900^\circ C$$

B
$$1000^\circ C$$

C
$$900^\circ C$$

D
$$1200^\circ C$$

Solution

$$At\quad 100℃,\quad 2M+{ O }_{ 2 }\longrightarrow 2MO;\Delta G=-921kJ/mole\\ \quad \quad \quad \quad \quad \quad \quad \therefore 2MO\longrightarrow 2M+{ O }_{ 2 };\Delta G=921kJ/mole\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad MO\longrightarrow M+\cfrac { 1 }{ 2 } { O }_{ 2 };\Delta G=460.5kJ/mole-------(i)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad 2C+{ O }_{ 2 }\longrightarrow 2CO;\Delta G=-432kJ/mole\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad C+\cfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow CO;\Delta G=-216kJ/mole--------(ii)\\ Adding\quad (i)\& (ii),\quad MO+C\xrightarrow { \quad \quad \quad \quad \Delta \quad \quad \quad } M+CO\uparrow ;\Delta G=460.5-216=244.5kJ/mole\\ \therefore \Delta G>0$$
$$\therefore $$ It is not feasible.
$$At\quad 1900℃,\quad 2M+{ O }_{ 2 }\longrightarrow 2MO;\Delta G=-300kJ/mole\\ \quad \quad \quad \quad \quad \quad \quad \therefore 2MO\longrightarrow 2M+{ O }_{ 2 };\Delta G=300kJ/mole\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad MO\longrightarrow M+\cfrac { 1 }{ 2 } { O }_{ 2 };\Delta G=150kJ/mole-------(iii)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad 2C+{ O }_{ 2 }\longrightarrow 2CO;\Delta G=-624kJ/mole\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad C+\cfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow CO;\Delta G=-312kJ/mole--------(iv)\\ Adding\quad (iii)\& (iv),\quad MO+C\xrightarrow { \quad \quad \quad \quad \Delta \quad \quad \quad } M+CO\uparrow ;\Delta G=150-312=-162kJ/mole\\ \therefore \Delta G<0$$
$$\therefore $$ It is feasible.
Question 10
1 / -0

Which of the following parameters does not charaterize the thermodynamic state of matter?

A
temperature

B
pressure

C
work

D
volume

Solution

From the above options Work depends on the shift from one state to another

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Thermodynamics Test - 35

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Thermodynamics Test - 35

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SHARING IS CARING

Question 1

Le Chatelier's law

van't Hoff's law

first law of thermodynamics

Hess's law.

Solution

Question 2

1169 J

769 J

1373 J

969 J

Solution

Question 3

spontaneous

non-spontaneous

at equilibrium

unpredictable

Solution

Question 4

temperature only

volume only

both volume and temperature

neither volume nor temperature

Solution

Question 5

$$T_A > T_C > T_B$$

$$T_A > T_B > T_C$$

$$T_B > T_C > T_A$$

$$T_C > T_B > T_A$$

Solution

Question 6

$$2.44\times { 10 }^{ 5 }$$J

$$1.62\times { 10 }^{ 5 }$$J

$$1.24\times { 10 }^{ 5 }$$J

$$2.62\times { 10 }^{ 5 }$$J

Solution

Question 7

30 J $${ s }^{ -1 }$$

40 J $${ s }^{ -1 }$$

50 J $${ s }^{ -1 }$$

60 J $${ s }^{ -1 }$$

Solution

Question 8

Same in I and II but not in III and IV

Same in III and IV but not in I and II

Same in I,II and III but not in IV

Same in all of the four cases

Solution

Question 9

$$1900^\circ C$$

$$1000^\circ C$$

$$900^\circ C$$

$$1200^\circ C$$

Solution

Question 10

temperature

pressure

work

volume

Solution

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