Q. For a reaction at equilibrium, the change in Gibbs free energy (ΔG) is equal to:
A.
ΔH - TΔS
B.
0
C.
ΔS - TΔH
D.
ΔH + TΔS
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Solution
At equilibrium, the change in Gibbs free energy (ΔG) is zero, indicating that the system is at maximum entropy.
Correct Answer:
B
— 0
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Q. For a reversible process, the change in entropy of the system is equal to the heat absorbed divided by the temperature. What is the formula?
A.
ΔS = Q/T
B.
ΔS = T/Q
C.
ΔS = Q*T
D.
ΔS = Q + T
Show solution
Solution
The change in entropy (ΔS) for a reversible process is given by ΔS = Q/T, where Q is the heat absorbed and T is the temperature.
Correct Answer:
A
— ΔS = Q/T
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Q. For a reversible process, the change in entropy of the system is equal to the heat absorbed divided by the temperature. This is expressed as:
A.
ΔS = Q/T
B.
ΔS = T/Q
C.
ΔS = Q + T
D.
ΔS = Q - T
Show solution
Solution
For a reversible process, the change in entropy (ΔS) is given by ΔS = Q/T, where Q is the heat absorbed and T is the temperature.
Correct Answer:
A
— ΔS = Q/T
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Q. For a reversible process, the change in entropy of the universe is:
A.
Zero
B.
Positive
C.
Negative
D.
Undefined
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Solution
For a reversible process, the change in entropy of the universe is zero, as the system and surroundings are in equilibrium.
Correct Answer:
A
— Zero
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Q. For a spontaneous process, the change in entropy of the universe must be:
A.
Zero
B.
Positive
C.
Negative
D.
Undefined
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Solution
For a spontaneous process, the total entropy change of the universe (system + surroundings) must be positive.
Correct Answer:
B
— Positive
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Q. For a spontaneous process, the change in Gibbs free energy (ΔG) is related to entropy (ΔS) how?
A.
ΔG = ΔH - TΔS
B.
ΔG = TΔS - ΔH
C.
ΔG = ΔS - ΔH
D.
ΔG = ΔH + TΔS
Show solution
Solution
The relationship is given by ΔG = ΔH - TΔS, where ΔG must be negative for a spontaneous process.
Correct Answer:
A
— ΔG = ΔH - TΔS
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Q. For a spontaneous process, the change in Gibbs free energy (ΔG) is related to entropy (ΔS) by which of the following equations?
A.
ΔG = ΔH + TΔS
B.
ΔG = ΔH - TΔS
C.
ΔG = TΔS - ΔH
D.
ΔG = ΔS - ΔH
Show solution
Solution
The correct relationship is ΔG = ΔH - TΔS, where ΔG must be negative for a spontaneous process.
Correct Answer:
B
— ΔG = ΔH - TΔS
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Q. For a spontaneous process, the change in Gibbs free energy (ΔG) is:
A.
Positive
B.
Negative
C.
Zero
D.
Undefined
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Solution
For a process to be spontaneous, the change in Gibbs free energy (ΔG) must be negative.
Correct Answer:
B
— Negative
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Q. If 100 J of heat is added to a system at a constant temperature of 300 K, what is the change in entropy?
A.
0.33 J/K
B.
0.25 J/K
C.
0.5 J/K
D.
0.75 J/K
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Solution
The change in entropy ΔS = Q/T = 100 J / 300 K = 0.33 J/K.
Correct Answer:
A
— 0.33 J/K
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Q. If the entropy of a system increases, what can be inferred about the spontaneity of the process?
A.
The process is non-spontaneous
B.
The process is spontaneous
C.
The process is at equilibrium
D.
None of the above
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Solution
An increase in entropy indicates that the process is spontaneous, as per the second law of thermodynamics.
Correct Answer:
B
— The process is spontaneous
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Q. If the temperature of a system is increased, what happens to its entropy?
A.
It decreases
B.
It remains constant
C.
It increases
D.
It becomes zero
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Solution
As the temperature of a system increases, the kinetic energy of the particles increases, leading to greater disorder and thus an increase in entropy.
Correct Answer:
C
— It increases
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Q. In a closed system, if the temperature increases, what happens to the entropy?
A.
It decreases
B.
It increases
C.
It remains constant
D.
It becomes zero
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Solution
In a closed system, an increase in temperature generally leads to an increase in entropy, as the molecular motion becomes more chaotic.
Correct Answer:
B
— It increases
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Q. In an irreversible process, the change in entropy of the universe is:
A.
Zero
B.
Positive
C.
Negative
D.
Undefined
Show solution
Solution
In an irreversible process, the change in entropy of the universe is positive, indicating that the total entropy increases.
Correct Answer:
B
— Positive
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Q. In which of the following reactions is the entropy change expected to be negative?
A.
N2(g) + 3H2(g) → 2NH3(g)
B.
C(s) + O2(g) → CO2(g)
C.
2H2(g) + O2(g) → 2H2O(g)
D.
CaCO3(s) → CaO(s) + CO2(g)
Show solution
Solution
The reaction N2(g) + 3H2(g) → 2NH3(g) results in a decrease in the number of gas molecules, leading to a negative change in entropy.
Correct Answer:
A
— N2(g) + 3H2(g) → 2NH3(g)
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Q. In which of the following scenarios is the entropy of the system likely to decrease?
A.
Ice melting
B.
Water freezing
C.
Gas expanding
D.
Liquid evaporating
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Solution
The entropy of the system decreases when water freezes, as the molecules become more ordered in the solid state.
Correct Answer:
B
— Water freezing
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Q. In which of the following scenarios would the entropy of the system decrease?
A.
Ice melting
B.
Water evaporating
C.
Gas compressing
D.
Sugar dissolving in water
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Solution
When a gas is compressed, the number of microstates decreases, leading to a decrease in entropy.
Correct Answer:
C
— Gas compressing
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Q. The entropy change for a phase transition at constant temperature is given by:
A.
ΔS = ΔH/T
B.
ΔS = T/ΔH
C.
ΔS = ΔH*T
D.
ΔS = ΔH + T
Show solution
Solution
For a phase transition at constant temperature, the change in entropy is given by ΔS = ΔH/T, where ΔH is the enthalpy change.
Correct Answer:
A
— ΔS = ΔH/T
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Q. The entropy change for a reaction can be calculated using which of the following?
A.
ΔS = ΣS(products) - ΣS(reactants)
B.
ΔS = ΣS(reactants) - ΣS(products)
C.
ΔS = Q/T
D.
ΔS = W/T
Show solution
Solution
The change in entropy for a reaction is calculated using the formula ΔS = ΣS(products) - ΣS(reactants).
Correct Answer:
A
— ΔS = ΣS(products) - ΣS(reactants)
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Q. The entropy of a perfect crystal at absolute zero is given by:
Show solution
Solution
According to the third law of thermodynamics, the entropy of a perfect crystal at absolute zero is zero.
Correct Answer:
B
— 0
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Q. The entropy of a perfect crystal at absolute zero is:
A.
Maximum
B.
Minimum
C.
Undefined
D.
Infinite
Show solution
Solution
According to the third law of thermodynamics, the entropy of a perfect crystal at absolute zero is zero, which is the minimum value.
Correct Answer:
B
— Minimum
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Q. The entropy of a perfect crystal at absolute zero temperature is given by which law?
A.
Third law of thermodynamics
B.
First law of thermodynamics
C.
Second law of thermodynamics
D.
Zeroth law of thermodynamics
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Solution
The third law of thermodynamics states that the entropy of a perfect crystal approaches zero as the temperature approaches absolute zero.
Correct Answer:
A
— Third law of thermodynamics
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Q. The entropy of a perfect crystal at absolute zero temperature is given by:
A.
0
B.
1
C.
Infinity
D.
Depends on the substance
Show solution
Solution
According to the third law of thermodynamics, the entropy of a perfect crystal at absolute zero is exactly zero.
Correct Answer:
A
— 0
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Q. The entropy of a perfect crystalline substance at absolute zero is given by which law?
A.
Third law of thermodynamics
B.
First law of thermodynamics
C.
Second law of thermodynamics
D.
Zeroth law of thermodynamics
Show solution
Solution
The third law of thermodynamics states that the entropy of a perfect crystalline substance approaches zero as the temperature approaches absolute zero.
Correct Answer:
A
— Third law of thermodynamics
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Q. The entropy of a perfect crystalline substance at absolute zero is:
A.
Zero
B.
Maximum
C.
Undefined
D.
Infinite
Show solution
Solution
According to the third law of thermodynamics, the entropy of a perfect crystalline substance at absolute zero is zero.
Correct Answer:
A
— Zero
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Q. What is the change in entropy when 1 mole of an ideal gas expands isothermally and reversibly from volume V1 to V2?
A.
R ln(V2/V1)
B.
R (V2 - V1)
C.
R (V1/V2)
D.
0
Show solution
Solution
The change in entropy for an isothermal and reversible expansion of an ideal gas is given by ΔS = nR ln(V2/V1). For 1 mole, n = 1, hence ΔS = R ln(V2/V1).
Correct Answer:
A
— R ln(V2/V1)
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Q. What is the change in entropy when 1 mole of an ideal gas expands isothermally from volume V1 to V2?
A.
R ln(V2/V1)
B.
R (V2 - V1)
C.
R (V1/V2)
D.
0
Show solution
Solution
The change in entropy for an isothermal expansion of an ideal gas is given by ΔS = nR ln(V2/V1). For 1 mole, it simplifies to ΔS = R ln(V2/V1).
Correct Answer:
A
— R ln(V2/V1)
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Q. What is the effect of increasing temperature on the entropy of a substance?
A.
Entropy decreases
B.
Entropy increases
C.
Entropy remains constant
D.
Entropy becomes zero
Show solution
Solution
Increasing temperature generally increases the kinetic energy of particles, leading to greater disorder and thus an increase in entropy.
Correct Answer:
B
— Entropy increases
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Q. What is the effect of temperature on the entropy of a substance?
A.
Entropy decreases with increasing temperature.
B.
Entropy increases with increasing temperature.
C.
Entropy remains constant with temperature.
D.
Temperature has no effect on entropy.
Show solution
Solution
As temperature increases, the kinetic energy of particles increases, leading to greater disorder and thus higher entropy.
Correct Answer:
B
— Entropy increases with increasing temperature.
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Q. What is the entropy change for a system that undergoes a phase transition at constant temperature?
A.
ΔS = 0
B.
ΔS = Q/T
C.
ΔS = T/Q
D.
ΔS = Q + T
Show solution
Solution
During a phase transition at constant temperature, the change in entropy is given by ΔS = Q/T, where Q is the heat absorbed or released.
Correct Answer:
B
— ΔS = Q/T
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Q. What is the entropy change for the isothermal expansion of an ideal gas from volume V1 to V2 at temperature T?
A.
R ln(V2/V1)
B.
R (V2 - V1)/T
C.
0
D.
R (V1/V2)
Show solution
Solution
The entropy change for an isothermal expansion is given by ΔS = nR ln(V2/V1). For 1 mole, ΔS = R ln(V2/V1).
Correct Answer:
A
— R ln(V2/V1)
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Showing 1 to 30 of 57 (2 Pages)
Entropy MCQ & Objective Questions
Understanding the concept of entropy is crucial for students preparing for various exams. It not only helps in grasping fundamental principles of thermodynamics but also enhances problem-solving skills. Practicing MCQs and objective questions on entropy can significantly improve your exam performance by familiarizing you with important questions and concepts that frequently appear in assessments.
What You Will Practise Here
Definition and significance of entropy in thermodynamics
Key formulas related to entropy calculations
Understanding the second law of thermodynamics
Entropy changes in reversible and irreversible processes
Diagrams illustrating entropy changes in various systems
Applications of entropy in real-world scenarios
Common misconceptions about entropy and its implications
Exam Relevance
The topic of entropy is frequently tested in CBSE, State Boards, NEET, and JEE exams. Students can expect questions that require them to apply the concept of entropy to solve problems or explain phenomena. Common question patterns include numerical problems, conceptual questions, and theoretical explanations, making it essential to be well-versed in both the definitions and applications of entropy.
Common Mistakes Students Make
Confusing entropy with energy; understanding their distinct roles is vital.
Misapplying the second law of thermodynamics in problem-solving.
Overlooking the significance of reversible vs. irreversible processes in calculations.
Failing to interpret graphical representations of entropy changes correctly.
FAQs
Question: What is entropy in simple terms?Answer: Entropy is a measure of the disorder or randomness in a system, reflecting how energy is distributed within that system.
Question: How can I improve my understanding of entropy for exams?Answer: Regular practice of entropy MCQ questions and reviewing key concepts can enhance your understanding and retention of the topic.
Ready to boost your exam preparation? Dive into our practice MCQs on entropy and solidify your understanding of this essential concept. The more you practice, the better you will score!
