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Thermodynamics and Kinetic Theory Questions for AGNIVEER VAYU

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Why this topic matters · 8 min read
Thermodynamics and Kinetic Theory appear in 8-12% of Agniveer Vayu physics papers, split between conceptual questions (laws of thermodynamics, heat engines) and numerical problems (ideal gas law, kinetic energy, work-heat calculations). Recent papers emphasize first law applications, efficiency calculations, and molecular speed distributions. Expect 2-3 questions mixing theory recall with quick calculations.

Kinetic Theory of Gases

Kinetic theory explains gas behavior by treating gas molecules as tiny, fast-moving particles in random motion. The pressure a gas exerts comes from billions of molecular collisions with container walls per second. Temperature is directly linked to the average kinetic energy of molecules—hotter gas means faster-moving particles. This theory bridges microscopic particle behavior with macroscopic properties we measure (pressure, volume, temperature).

  • Pressure arises from momentum transfer during molecular collisions with walls
  • Average kinetic energy per molecule = (3/2)kT, where k is Boltzmann constant and T is absolute temperature
  • All gas molecules at same temperature have same average kinetic energy, regardless of molecular mass
  • Root mean square speed (RMS) increases with temperature, decreases with molecular mass
  • Ideal gas assumption: molecules occupy negligible volume, no intermolecular forces except collisions
Key formulas
RMS Speed
v_rms = sqrt(3kT/m) = sqrt(3RT/M)
When: Finding typical molecular speed; k is Boltzmann constant, m is single molecule mass, M is molar mass
Ideal Gas Law
PV = nRT or PV = NkT
When: Relating pressure, volume, temperature; n is moles, N is number of molecules
Average Kinetic Energy
KE_avg = (3/2)kT per molecule or (3/2)nRT per mole
When: Connecting temperature to molecular motion energy
Worked examples

A container has nitrogen gas at 300K. Find RMS speed. M(N2)=28 g/mol, R=8.314 J/(mol·K). v_rms = sqrt(3×8.314×300/0.028) = 517 m/s.

Two gases at same temperature: oxygen (M=32) and hydrogen (M=2). Hydrogen molecules move faster by factor sqrt(32/2) = 4 times faster RMS speed.

First Law of Thermodynamics

The first law states that energy cannot be created or destroyed—it only changes form. When heat Q is added to a system, it either increases internal energy U (raising temperature) or does work W on surroundings (expanding, lifting, etc.). This is the foundation of all heat engine problems. Think of it as an energy balance sheet: money in (heat) = money stored (internal energy) + money spent (work).

  • First Law: dU = Q - W, where Q is heat added to system, W is work done by system
  • If Q > 0: heat flows into system; if Q < 0: heat flows out
  • If W > 0: system does work on surroundings (expansion); if W < 0: surroundings do work on system (compression)
  • For ideal gas, internal energy depends only on temperature: U = nC_v T
  • Adiabatic process (Q=0): all work comes from internal energy change, system cools as it expands
Key formulas
First Law
dU = Q - W or Q = dU + W
When: Any thermodynamic process; track energy flow
Work by Gas
W = integral(P dV) or W = nRT ln(V_f/V_i) for isothermal
When: Calculating work in expansion/compression
Internal Energy Change
dU = nC_v dT
When: Relating temperature change to energy; C_v = (f/2)R where f is degrees of freedom
Worked examples

A gas expands isothermally (constant T) from 1L to 2L at 300K, n=1 mol. dU=0 (T constant). Q = W = nRT ln(2) = 8.314×300×0.693 = 1730 J. All heat goes to work.

Gas compressed adiabatically (Q=0). Work done on gas = 500 J. Then dU = -(-500) = 500 J. Temperature rises.

Second Law and Heat Engines

The second law states that heat naturally flows from hot to cold, never spontaneously the other way. For heat engines (like car engines), this means you cannot convert all input heat into useful work—some must be rejected as waste heat to a cold reservoir. Efficiency measures what fraction of input heat becomes useful work. Real engines always have efficiency less than the theoretical Carnot limit.

  • Heat engine: absorbs heat Q_h from hot reservoir, does work W, rejects heat Q_c to cold reservoir
  • Energy balance: Q_h = W + Q_c (first law applied to cycle)
  • Efficiency: e = W/Q_h = 1 - Q_c/Q_h (always less than 100%)
  • Carnot efficiency (theoretical maximum): e_Carnot = 1 - T_c/T_h (depends only on absolute temperatures)
  • Real engines have lower efficiency due to friction, heat loss, irreversible processes
Key formulas
Engine Efficiency
e = W/Q_h = (Q_h - Q_c)/Q_h = 1 - Q_c/Q_h
When: Finding what fraction of input heat becomes useful work
Carnot Efficiency
e_Carnot = 1 - T_c/T_h
When: Upper limit for any heat engine; T in Kelvin
Coefficient of Performance (Refrigerator)
COP = Q_c/W = T_c/(T_h - T_c)
When: Measuring refrigerator/AC efficiency (inverse of engine)
Worked examples

Heat engine absorbs 1000 J from hot reservoir, rejects 600 J to cold. Work output W = 1000 - 600 = 400 J. Efficiency e = 400/1000 = 40%.

Carnot engine between T_h = 600K and T_c = 300K. Max efficiency = 1 - 300/600 = 50%. Real engine might achieve 35-40%.

Specific Heat and Degrees of Freedom

Specific heat capacity tells you how much energy is needed to raise temperature of a substance by 1 degree. For gases, it depends on whether volume or pressure is held constant. A gas at constant volume (C_v) heats up faster than at constant pressure (C_p) because at constant pressure some energy goes into expansion work. The difference C_p - C_v = R comes directly from the first law.

  • C_v = heat capacity at constant volume (all heat raises internal energy)
  • C_p = heat capacity at constant pressure (heat raises internal energy and does expansion work)
  • For ideal gas: C_p - C_v = R (per mole)
  • C_v = (f/2)R where f is degrees of freedom: f=3 (monatomic), f=5 (diatomic), f=6 (polyatomic)
  • Ratio gamma = C_p/C_v: monatomic 5/3, diatomic 7/5, polyatomic 9/7
Key formulas
Mayer's Relation
C_p - C_v = R
When: Relating heat capacities at constant pressure and volume
Heat Capacity from Degrees of Freedom
C_v = (f/2)R, C_p = ((f+2)/2)R
When: Finding C_v, C_p for ideal gas; f = translational + rotational + vibrational degrees
Heat Required
Q = nC_v dT (constant volume) or Q = nC_p dT (constant pressure)
When: Calculating heat absorbed/released in temperature change
Worked examples

Diatomic gas (f=5): C_v = (5/2)R = 20.8 J/(mol·K), C_p = (7/2)R = 29.1 J/(mol·K). Ratio gamma = 7/5 = 1.4.

1 mole of monatomic gas heated at constant volume from 300K to 400K. Q = nC_v dT = 1×(3/2)×8.314×100 = 1247 J.

Isothermal, Adiabatic, and Cyclic Processes

Different thermodynamic processes follow different paths on a P-V diagram. Isothermal (constant T) means internal energy stays fixed, so all heat becomes work. Adiabatic (no heat exchange) means the system is thermally isolated—work comes entirely from internal energy, causing temperature to drop during expansion. Cyclic processes return to starting state; net work equals area enclosed on P-V diagram.

  • Isothermal: T constant, dU=0, Q=W, follows hyperbola PV=constant on P-V diagram
  • Adiabatic: Q=0, dU=-W, temperature drops during expansion, follows steeper curve PV^gamma=constant
  • Isobaric: P constant, W=P(V_f-V_i), Q=nC_p dT
  • Isochoric: V constant, W=0, Q=nC_v dT, all heat changes internal energy
  • Cyclic: returns to initial state, dU_cycle=0, W_cycle=Q_cycle (net work = net heat)
Key formulas
Isothermal Work
W = nRT ln(V_f/V_i) = nRT ln(P_i/P_f)
When: Work done in isothermal expansion/compression
Adiabatic Relation
TV^(gamma-1) = constant or P^(1-gamma) T^gamma = constant
When: Relating T, V, P in adiabatic process without calculating work explicitly
Isobaric Work
W = P(V_f - V_i) = nR(T_f - T_i)
When: Work at constant pressure
Worked examples

Isothermal expansion of 2 mol ideal gas at 400K from 10L to 20L. W = 2×8.314×400×ln(2) = 4608 J. Q = W = 4608 J (no temperature change).

Adiabatic compression of diatomic gas (gamma=1.4) from 300K. If volume halves, T_f = T_i × (V_i/V_f)^(gamma-1) = 300×2^0.4 = 476K.

⚠ Common mistakes to avoid
  • Confusing sign convention: W is work done BY system (positive during expansion). Many students reverse this, causing sign errors in first law.
  • Using Celsius instead of Kelvin in gas laws and Carnot efficiency. Temperature must always be absolute (Kelvin) in thermodynamic formulas.
  • Assuming all heat added increases temperature. In isothermal processes, heat does work without raising T. In phase changes, heat changes state without raising T.
  • Forgetting that Carnot efficiency is a ceiling, not typical. Real engines achieve 30-50% of Carnot limit. Exam often asks 'why is actual efficiency lower'—answer is irreversibility.
  • Mixing up C_v and C_p. Remember: C_p is always larger (extra energy for expansion work). For monatomic, C_v=(3/2)R; for diatomic, C_v=(5/2)R.
🧠 Memory aids
  • FIRST LAW: dU = Q - W. Think 'Change in energy = Heat in - Work out'. Like bank account: deposit (Q) minus withdrawal (W) = net change (dU).
  • CARNOT LIMIT: e = 1 - T_cold/T_hot. Higher temperature difference = higher efficiency. Never reaches 100% because T_cold can't be zero Kelvin.
  • RMS SPEED: v = sqrt(3RT/M). Hotter gas = faster molecules. Heavier molecules = slower. Nitrogen faster than oxygen at same T because... wait, no: oxygen heavier so slower. Mnemonic: 'Hot and Light = Fast'.
  • DEGREES OF FREEDOM: Monatomic (3D motion only) = 3. Diatomic (3D + 2 rotations) = 5. Polyatomic (3D + 3 rotations) = 6. Add vibrational at very high T.
  • ADIABATIC vs ISOTHERMAL: Adiabatic is like a thermos (no heat escape)—temperature changes. Isothermal is like a pot on stove (heat flows in/out)—temperature constant.
🎯 AGNIVEER VAYU exam tips
  • Agniveer Vayu papers typically include 1-2 straightforward first law problems: given Q and W, find dU, or vice versa. These are quick points if you remember sign convention.
  • Heat engine efficiency questions are common. Expect: 'Engine absorbs X joules, rejects Y joules. Find efficiency.' Also: 'Compare actual vs Carnot efficiency'—test conceptual understanding.
  • Kinetic theory questions often ask for RMS speed or relate temperature to molecular energy. Usually one numerical problem. Watch units: molar mass in kg/mol, not g/mol.
  • Adiabatic process questions sometimes appear as 'gas expands without heat exchange.' Key insight: dU = -W, so temperature drops. Students often forget temperature changes in adiabatic.
  • Recent papers (2022-2024) show increased focus on cyclic processes and P-V diagrams. Expect one question asking to identify process type or calculate net work from diagram. Practice sketching isothermal vs adiabatic curves.

Sample questions

Q1 · medium · AI-verified
Which of the following processes in a thermodynamic cycle has ΔU = 0?
  1. Adiabatic process
  2. Isochoric process
  3. Isothermal process
  4. Isobaric process
Q2 · medium · AI-verified
The equation of state for n moles of an ideal gas is PV = nRT. If both the pressure and volume of an ideal gas are doubled, its absolute temperature becomes:
  1. 8 times the original
  2. Unchanged
  3. 2 times the original
  4. 4 times the original
Q3 · hard · AI-verified
In a Carnot engine, the source temperature is 527°C and the sink temperature is 27°C. If the engine absorbs 10,000 J of heat per cycle, what is the work done per cycle?
  1. 6250 J
  2. 7500 J
  3. 3750 J
  4. 5000 J
Q4 · easy · AI-verified
In an isothermal process, which thermodynamic quantity remains constant?
  1. Internal energy change is non-zero
  2. Pressure
  3. Temperature
  4. Volume
Q5 · easy · AI-verified
According to the first law of thermodynamics, which of the following is conserved?
  1. Pressure
  2. Energy
  3. Temperature
  4. Entropy
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