Relevant Titles
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First Law of Thermodynamics & Internal Energy — Class 11 Chemistry MCQs
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CBSE Class 11 Thermodynamics Quiz: First Law Practice Questions
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NCERT-Aligned MCQs on ΔU = q + w for Class 11 Chemistry
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30-Question Timed Quiz: First Law of Thermodynamics (Class 11)
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CBSE Class 11 Chemistry Practice — Internal Energy and Heat/Work MCQs
Introduction
Mastering the First Law of Thermodynamics and internal energy is essential for CBSE Class 11 students studying Chapter 6: Thermodynamics. These carefully crafted MCQs focus on the conservation of energy (ΔU = q + w), sign conventions for heat and work, and common process types such as isochoric, isobaric, isothermal and adiabatic transformations. The set also reinforces concepts connecting internal energy to temperature changes and heat capacities, along with practical problem-solving involving calorimetry and PV work. Each question is aligned strictly with the NCERT syllabus to ensure exam relevance and clarity. Instant explanations after submission help you learn from mistakes and solidify concepts. This 30-question, 30-minute timed quiz is ideal for focused revision: practice unit conversions, energy calculations, and interpretation of thermodynamic scenarios under exam conditions. Attempting this test will sharpen your numerical skills, deepen conceptual understanding, and boost confidence for the CBSE board exam and competitive tests and application.
Sample MCQs (with answers & explanations)
Q1. The First Law of Thermodynamics is best described as which principle?
A) Energy may be created or destroyed
B) Energy is conserved; ΔU = q + w
C) Entropy always increases
D) Mass is conserved only
Answer: B
Explanation: The First Law states that energy cannot be created or destroyed; the change in internal energy (ΔU) equals heat added to the system (q) plus work done on the system (w), sign convention dependent.
Q2. A system absorbs 200 J of heat and does 75 J of work on the surroundings. What is ΔU (using ΔU = q + w with work done by system negative)?
A) +275 J
B) +125 J
C) −125 J
D) −275 J
Answer: B (+125 J)
Explanation: With the sign convention w (work done on system) = −75 J (because system does work on surroundings). So ΔU = q + w = +200 + (−75) = +125 J.
Q3. Which quantity is a state function (depends only on initial and final states)?
A) Heat (q)
B) Work (w)
C) Internal energy (U)
D) Path taken by process
Answer: C
Explanation: Internal energy (U) is a state function — it depends only on the system’s state, not on how the change occurred. Heat and work are path functions.
Q4. For an ideal gas undergoing an isothermal reversible expansion, which statement is correct?
A) ΔU > 0
B) ΔU < 0
C) ΔU = 0 and q = −w
D) q = 0 and w = 0
Answer: C
Explanation: For an ideal gas, internal energy depends only on temperature; in an isothermal process ΔT = 0 ⇒ ΔU = 0. Therefore heat absorbed equals negative of work done by the system (q = −w).
Q5. A gas expands against a constant external pressure of 2.0 atm from 1.0 L to 3.0 L. What is the work done (in J)? (1 L·atm = 101.325 J)
A) +40.5 J
B) −40.5 J
C) +405 J
D) −405 J
Answer: D (−405 J)
Explanation: w = −P_ext ΔV = −2.0 atm × (3.0 − 1.0) L = −4.0 L·atm = −4.0 × 101.325 J ≈ −405 J. Negative sign indicates work done by the system.