Physics Lab

Unit 8: Thermodynamics and Kinetic Theory

This is a 3–4 MCQ-per-year unit on NEET. It is content-rich but formula-heavy. The dominant question types are: (i) calorimetry mixing problems, (ii) thermal conduction in series/parallel rods, (iii) kinetic-theory ratios involving RMS, average, most probable speeds, (iv) first-law calculations on standard processes (isothermal, adiabatic), and (v) Carnot efficiency.

Concept Map

  • Heat and Temperature scales
  • Thermal expansion — linear, area, volume
  • Calorimetry — specific heat, latent heat
  • Heat transfer — conduction (Fourier), convection (qualitative), radiation (Stefan, Wien, Newton's cooling)
  • Kinetic Theory of Gases — postulates, pressure formula, kT, RMS, equipartition, degrees of freedom, CpC_p, CvC_v, γ\gamma, mean free path
  • First Law of Thermodynamics and processes (isothermal, adiabatic, isobaric, isochoric)
  • Cyclic processes, heat engines, refrigerator
  • Carnot cycle and second law

Topic 1: Temperature Scales and Thermal Expansion

Sub-topic A: Temperature Scales

TC100=TF32180=TK273.15100,TK=TC+273.15.\frac{T_C}{100} = \frac{T_F - 32}{180} = \frac{T_K - 273.15}{100}, \quad T_K = T_C + 273.15.

Absolute zero: TK=0T_K = 0 K. All thermodynamic temperatures use Kelvin.

Sub-topic B: Thermal Expansion

TypeRelationCoefficient
LinearL=L0(1+αΔT)L = L_0(1 + \alpha\Delta T)α\alpha
AreaA=A0(1+βΔT)A = A_0(1 + \beta\Delta T)β=2α\beta = 2\alpha
VolumeV=V0(1+γΔT)V = V_0(1 + \gamma\Delta T)γ=3α\gamma = 3\alpha

For an isotropic solid, α:β:γ=1:2:3\alpha : \beta : \gamma = 1 : 2 : 3.

Anomalous expansion of water: water has maximum density at 4 °C; below this, expansion occurs as TT decreases. This is why lakes freeze top-down.

Topic 2: Calorimetry

Sub-topic A: Specific Heat and Heat Capacity

Heat needed to change temperature of mass mm by ΔT\Delta T:

Q=msΔT,Q = m s \Delta T,

where ss is specific heat capacity (J/(kg·K)). For water, sw4186 J/(kg K)=1 cal/(g °C)s_w \approx 4186\ \text{J/(kg K)} = 1\ \text{cal/(g °C)}.

Heat capacity C=msC = ms.

Sub-topic B: Latent Heat

For a phase change at constant temperature:

Q=mL,Q = m L,

with LfusionL_\text{fusion} (solid ↔ liquid) and LvapL_\text{vap} (liquid ↔ vapour). For water: Lf=80 cal/gL_f = 80\ \text{cal/g}, Lv=540 cal/gL_v = 540\ \text{cal/g}.

Sub-topic C: Mixing (Principle of Calorimetry)

Heat lost by hot body = heat gained by cold body (no external heat loss). Solve for final TT:

m1s1(T1T)=m2s2(TT2).m_1 s_1 (T_1 - T) = m_2 s_2 (T - T_2).

For three substances: balance net heat.

Topic 3: Heat Transfer

Sub-topic A: Conduction — Fourier's Law

Rate of heat flow through a rod of cross-section AA, length LL, conductivity KK, temperature difference ΔT\Delta T:

dQdt=KAΔTL.\frac{dQ}{dt} = K A \frac{\Delta T}{L}.

Thermal resistance: R=L/(KA)R = L/(KA). So dQ/dt=ΔT/RdQ/dt = \Delta T/R — Ohm's-law analogy.

Series rods: Req=R1+R2R_\text{eq} = R_1 + R_2. Parallel rods: 1/Req=1/R1+1/R21/R_\text{eq} = 1/R_1 + 1/R_2.

Sub-topic B: Convection

Heat transfer by bulk motion of fluid. Two types:

  • Natural convection: density-driven (warm air rises).
  • Forced convection: external pumping (fan).

Quantitatively, dQ/dt=hAΔTdQ/dt = h A \Delta T (Newton).

Sub-topic C: Radiation

All bodies emit thermal radiation. For a black body:

Stefan-Boltzmann law:

P=σAT4,σ=5.67×108 W/(m2K4).P = \sigma A T^4, \quad \sigma = 5.67 \times 10^{-8}\ \text{W/(m}^2\text{K}^4).

For a body with emissivity ϵ\epsilon: P=ϵσAT4P = \epsilon \sigma A T^4.

Net radiation in surroundings at T0T_0:

dQdt=ϵσA(T4T04).\frac{dQ}{dt} = \epsilon \sigma A (T^4 - T_0^4).

Wien's displacement law: peak emission wavelength

λmT=b,b=2.898×103 m K.\lambda_m T = b, \quad b = 2.898 \times 10^{-3}\ \text{m K}.

Hotter bodies emit at shorter wavelengths.

Newton's law of cooling (for small ΔT\Delta T):

dTdt=k(TT0).\frac{dT}{dt} = -k(T - T_0).

This is the linearised version of Stefan's law. Solution: T(t)=T0+(TiT0)ektT(t) = T_0 + (T_i - T_0) e^{-kt}.

Topic 4: Kinetic Theory of Gases

Sub-topic A: Postulates

  • Gas consists of identical molecules in random motion.
  • Molecules are point-like (volume negligible compared to container).
  • No intermolecular forces except during elastic collisions.
  • Collisions with walls are elastic and instantaneous.
  • Velocity distribution obeys Maxwell-Boltzmann.

Sub-topic B: Pressure of an Ideal Gas

P=13ρv2=13nmv2,P = \tfrac{1}{3} \rho \overline{v^2} = \tfrac{1}{3} n m \overline{v^2},

where nn is number density and v2\overline{v^2} is mean-square speed.

Sub-topic C: Temperature and Kinetic Energy

For monatomic ideal gas, average translational KE per molecule:

K=32kBT,\langle K \rangle = \tfrac{3}{2} k_B T,

with Boltzmann's constant kB=1.38×1023 J/Kk_B = 1.38 \times 10^{-23}\ \text{J/K}.

For nn moles, total translational KE is 32nRT\tfrac{3}{2} n R T, where R=NAkB=8.314 J/(mol K)R = N_A k_B = 8.314\ \text{J/(mol K)}.

Sub-topic D: Speed Distributions

For Maxwell-Boltzmann:

SpeedFormula
Most probablevp=2kBT/mv_p = \sqrt{2 k_B T/m}
Averagevˉ=8kBT/(πm)\bar v = \sqrt{8 k_B T/(\pi m)}
RMSvrms=3kBT/mv_\text{rms} = \sqrt{3 k_B T/m}

Ratio: vp:vˉ:vrms=2:8/π:31:1.128:1.224v_p : \bar v : v_\text{rms} = \sqrt{2} : \sqrt{8/\pi} : \sqrt{3} \approx 1 : 1.128 : 1.224.

So vrms>vˉ>vpv_\text{rms} > \bar v > v_p.

Sub-topic E: Equipartition Theorem

Each quadratic degree of freedom contributes 12kBT\tfrac{1}{2} k_B T to the average energy.

GasDoF (per molecule)CvC_v per moleCpC_pγ=Cp/Cv\gamma = C_p/C_v
Monatomic (He, Ne, Ar)3 (translational)32R\tfrac{3}{2}R52R\tfrac{5}{2}R5/31.675/3 \approx 1.67
Diatomic (N₂, O₂) at moderate T5 (3 trans + 2 rot)52R\tfrac{5}{2}R72R\tfrac{7}{2}R7/5=1.47/5 = 1.4
Diatomic (vib. excited, high T)772R\tfrac{7}{2}R92R\tfrac{9}{2}R9/71.299/7 \approx 1.29
Triatomic (linear)772R\tfrac{7}{2}R92R\tfrac{9}{2}R9/79/7
Triatomic (non-linear)6 (3 trans + 3 rot)3R3R4R4R4/31.334/3 \approx 1.33

Mayer's relation: CpCv=RC_p - C_v = R.

Sub-topic F: Mean Free Path

λ=12πd2n,n=N/V.\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}, \quad n = N/V.

At STP, λ107 m\lambda \approx 10^{-7}\ \text{m}.

Topic 5: First Law of Thermodynamics

Sub-topic A: Statement

ΔU=QW,\Delta U = Q - W,

where WW is work done BY the gas (some books use Q=ΔU+WQ = \Delta U + W). Be consistent with sign convention.

Internal energy UU is a state function; QQ and WW are path functions.

For an ideal gas, U=U(T)U = U(T) only (depends only on temperature, not on VV or PP).

Sub-topic B: Work Done by a Gas

W=PdV.W = \int P\,dV.

Sign: W>0W > 0 for expansion, W<0W < 0 for compression.

Sub-topic C: Specific Processes

ProcessDefinitionWorkHeatΔU\Delta U
IsothermalTT constW=nRTln(V2/V1)W = n R T \ln(V_2/V_1)Q=WQ = W0
AdiabaticQ=0Q = 0W=(P1V1P2V2)/(γ1)=ΔUW = (P_1 V_1 - P_2 V_2)/(\gamma - 1) = -\Delta U0W-W
IsobaricPP constW=PΔV=nRΔTW = P\Delta V = nR\Delta TQ=nCpΔTQ = nC_p\Delta TnCvΔTnC_v\Delta T
IsochoricVV const0Q=nCvΔTQ = nC_v\Delta TQQ

Sub-topic D: Adiabatic Relations

For a reversible adiabatic of ideal gas:

PVγ=const,TVγ1=const,TγP1γ=const.P V^\gamma = \text{const}, \quad T V^{\gamma - 1} = \text{const}, \quad T^\gamma P^{1-\gamma} = \text{const}.

Adiabatic curve is steeper than isothermal on a PV diagram.

Sub-topic E: Cyclic Processes

For a cycle: ΔU=0\Delta U = 0, so Q=WQ = W. Net work = area enclosed by the cycle on the PV diagram. Sign: positive if traversed clockwise (heat engine), negative if counter-clockwise (refrigerator).

Topic 6: Heat Engines and the Second Law

Sub-topic A: Heat Engine

Operates between hot reservoir at THT_H and cold reservoir at TCT_C. Per cycle: absorbs QHQ_H, does work WW, rejects QCQ_C.

Efficiency:

η=WQH=1QCQH.\eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H}.

Sub-topic B: Refrigerator / Heat Pump

Operates in reverse — work input WW extracts QCQ_C from cold reservoir and dumps QH=QC+WQ_H = Q_C + W into hot reservoir.

Coefficient of performance (refrigerator):

βref=QC/W.\beta_\text{ref} = Q_C/W.

For a heat pump: βHP=QH/W=βref+1\beta_\text{HP} = Q_H/W = \beta_\text{ref} + 1.

Sub-topic C: Carnot Cycle

Reversible cycle: two isotherms (at THT_H and TCT_C) + two adiabats. Achieves maximum efficiency between given reservoirs:

ηCarnot=1TCTH.\eta_\text{Carnot} = 1 - \frac{T_C}{T_H}.

Carnot refrigerator: β=TC/(THTC)\beta = T_C/(T_H - T_C).

No engine can exceed Carnot efficiency (second law). For 100% efficiency we'd need TC=0T_C = 0 — impossible.

Sub-topic D: Second Law

Several equivalent statements:

  • Kelvin-Planck: No process can convert heat entirely into work in a cycle.
  • Clausius: Heat cannot flow spontaneously from a colder to a hotter body.
  • Entropy of an isolated system never decreases.

NEET Pattern MCQ Tips

  • Calorimetry mixing: ice + water + warmer water, find final TT.
  • Conduction through composite: series/parallel rods.
  • Stefan / Newton cooling: PT4P \propto T^4 or dT/dt=k(TT0)dT/dt = -k(T - T_0).
  • Kinetic theory: ratios of vp,vˉ,vrmsv_p, \bar v, v_\text{rms}; effect of doubling TT on vrmsv_\text{rms} (2\sqrt{2} times).
  • γ\gamma for ideal gas: 5/3,7/5,9/75/3, 7/5, 9/7.
  • Carnot efficiency: insert THT_H and TCT_C in kelvin.
  • Adiabatic vs isothermal: distinguish work formulas, slope comparisons.
  • Assertion-Reason: any heat engine efficiency << Carnot.

Common Confusions and Traps

  • γ=7/5\gamma = 7/5 for diatomic gas at room temperature (5 DoF, vibration frozen).
  • CpCv=RC_p - C_v = R per mole, not per molecule (per molecule it would be kBk_B).
  • For monatomic gas Cv=(3/2)RC_v = (3/2)R — students sometimes use the diatomic value.
  • In adiabatic, ΔU=W\Delta U = -W (work done by gas equals negative of internal energy change).
  • PV=nRTPV = nRT: R=8.314 J/(mol K)R = 8.314\ \text{J/(mol K)}; or use the per-molecule kBTk_B T.
  • Stefan's law: rate T4\propto T^4 in kelvin, never centigrade.
  • For Newton's cooling, the rate dT/dtdT/dt is proportional to (TT0)(T - T_0) in either kelvin or celsius (only differences matter).
  • A heat engine never has η=1\eta = 1 — second law.
  • The most probable speed is the smallest of vp,vˉ,vrmsv_p, \bar v, v_\text{rms}.

Quick Revision Card

  • Linear: β=2α\beta = 2\alpha; volume: γvol=3α\gamma_\text{vol} = 3\alpha.
  • Q=msΔTQ = ms\Delta T, Q=mLQ = mL.
  • Conduction: dQ/dt=KAΔT/LdQ/dt = KA\Delta T/L, R=L/(KA)R = L/(KA).
  • Stefan: P=ϵσAT4P = \epsilon\sigma A T^4; Wien: λmT=b\lambda_m T = b.
  • Newton's cooling: dT/dt=k(TT0)dT/dt = -k(T - T_0).
  • Pressure: P=13ρv2P = \tfrac{1}{3}\rho\overline{v^2}.
  • Avg KE per molecule (monatomic) =32kBT= \tfrac{3}{2}k_BT.
  • Speeds: vp=2kBT/mv_p = \sqrt{2k_BT/m}, vˉ=8kBT/πm\bar v = \sqrt{8k_BT/\pi m}, vrms=3kBT/mv_\text{rms} = \sqrt{3k_BT/m}.
  • Mayer: CpCv=RC_p - C_v = R; γ=Cp/Cv\gamma = C_p/C_v.
  • Monatomic: γ=5/3\gamma = 5/3; diatomic: γ=7/5\gamma = 7/5.
  • Adiabatic: PVγ=PV^\gamma = const.
  • Carnot: η=1TC/TH\eta = 1 - T_C/T_H; βref=TC/(THTC)\beta_\text{ref} = T_C/(T_H - T_C).
  • First law: ΔU=QW\Delta U = Q - W (with WW = work by gas).

Worked NEET Examples

Example 1: Mixing Hot and Cold Water

200 g water at 80 °C mixed with 100 g at 20 °C. Final temperature:

200×(80T)=100×(T20).200 \times (80 - T) = 100 \times (T - 20).

200×80200T=100T2000200 \times 80 - 200T = 100T - 2000, 18000=300T18000 = 300T, T=60T = 60 °C.

Example 2: Conduction Through Composite Rod

Two rods of equal length and cross-section but conductivities K1K_1 and K2K_2 in series:

Heat current must be equal in steady state. If hot end at T1T_1, cold at T2T_2, junction at TjT_j:

K1A(T1Tj)/L=K2A(TjT2)/L    Tj=(K1T1+K2T2)/(K1+K2).K_1 A (T_1 - T_j)/L = K_2 A (T_j - T_2)/L \implies T_j = (K_1 T_1 + K_2 T_2)/(K_1 + K_2).

Effective conductivity for series: Keff=2K1K2/(K1+K2)K_\text{eff} = 2 K_1 K_2/(K_1 + K_2) (harmonic-like mean).

For parallel arrangement (rods side by side, same length, different areas): Keff=(K1A1+K2A2)/(A1+A2)K_\text{eff} = (K_1 A_1 + K_2 A_2)/(A_1 + A_2).

Example 3: Black Body Radiation

A black body at 500 K emits at rate P0P_0. At 1000 K it emits at P=P0(1000/500)4=16P0P = P_0 (1000/500)^4 = 16 P_0.

Example 4: Carnot Efficiency

Hot reservoir at 400 K, cold at 300 K. Carnot efficiency:

η=1300/400=0.25=25%.\eta = 1 - 300/400 = 0.25 = 25\%.

So per 4 J absorbed at hot, 1 J becomes work, 3 J rejected to cold.

Example 5: RMS Speed of Oxygen

For O₂ at T = 300 K, molecular mass M=32M = 32 g/mol = 0.032 kg/mol:

vrms=3RT/M=3×8.314×300/0.0322.34×105484 m/s.v_\text{rms} = \sqrt{3RT/M} = \sqrt{3 \times 8.314 \times 300/0.032} \approx \sqrt{2.34 \times 10^5} \approx 484\ \text{m/s}.

Derivations

Pressure of an Ideal Gas

Consider a cube of side LL containing NN molecules each of mass mm. A molecule with vxv_x component impacts a face, transferring momentum 2mvx2 m v_x. Time between impacts on the same face: 2L/vx2L/v_x. So average force from one molecule: 2mvx/(2L/vx)=mvx2/L2 m v_x/(2L/v_x) = m v_x^2/L.

For NN molecules: total force = Nmvx2/LN m \overline{v_x^2}/L. Pressure on face of area L2L^2:

P=Nmvx2/L3=(N/V)mvx2.P = N m \overline{v_x^2}/L^3 = (N/V) m \overline{v_x^2}.

By isotropy, vx2=v2/3\overline{v_x^2} = \overline{v^2}/3:

P=(1/3)(N/V)mv2=(1/3)ρv2.P = (1/3) (N/V) m \overline{v^2} = (1/3)\rho \overline{v^2}.

Connection to Temperature

Compare with ideal gas law PV=nRT=NkBTPV = nRT = N k_B T:

v2=3kBT/m,12mv2=(3/2)kBT.\overline{v^2} = 3 k_B T/m, \quad \tfrac{1}{2} m \overline{v^2} = (3/2) k_B T.

So average translational KE per molecule = (3/2)kBT(3/2) k_B T.

Adiabatic Process for Ideal Gas

First law: dU=dWdU = -dW (since dQ=0dQ = 0). For ideal gas dU=nCvdTdU = n C_v dT and dW=PdVdW = P dV.

nCvdT+PdV=0.n C_v dT + P dV = 0.

From PV=nRTPV = nRT: PdV+VdP=nRdTP dV + V dP = nR dT, so dT=(PdV+VdP)/(nR)dT = (P dV + V dP)/(nR).

Substituting: Cv(PdV+VdP)/R+PdV=0C_v (P dV + V dP)/R + P dV = 0, so CvPdV+CvVdP+RPdV=0C_v P dV + C_v V dP + R P dV = 0, i.e. (Cv+R)PdV+CvVdP=0(C_v + R) P dV + C_v V dP = 0, or CpPdV+CvVdP=0C_p P dV + C_v V dP = 0.

Rearranging: γdV/V+dP/P=0\gamma\,dV/V + dP/P = 0, giving PVγ=PV^\gamma = constant.

Carnot Efficiency

Carnot cycle has two isothermals (at TH,TCT_H, T_C) and two adiabats. Heat absorbed at THT_H: QH=nRTHln(V2/V1)Q_H = nRT_H\ln(V_2/V_1). Heat rejected at TCT_C: QC=nRTCln(V3/V4)Q_C = nRT_C\ln(V_3/V_4). From adiabatic legs: THV2γ1=TCV3γ1T_H V_2^{\gamma-1} = T_C V_3^{\gamma-1} and THV1γ1=TCV4γ1T_H V_1^{\gamma-1} = T_C V_4^{\gamma-1}, so V2/V1=V3/V4V_2/V_1 = V_3/V_4. Therefore QC/QH=TC/THQ_C/Q_H = T_C/T_H and

η=1TC/TH.\eta = 1 - T_C/T_H.

Indicator Diagrams and Cyclic Work

For any cyclic process, work done by the gas equals the area enclosed on the P-V diagram. Direction:

  • Clockwise traversal: positive work (heat engine).
  • Counter-clockwise: negative work (refrigerator).

In a rectangular cycle on P-V plane: ABCDA with corners (V1,P1),(V2,P1),(V2,P2),(V1,P2)(V_1, P_1), (V_2, P_1), (V_2, P_2), (V_1, P_2). Work = (V2V1)(P1P2)(V_2 - V_1)(P_1 - P_2).

Two Specific Heat Relations

Mayer's Relation Derivation

For ideal gas, U=nCvTU = nC_v T depends only on T. At constant pressure: Q=nCpΔTQ = nC_p \Delta T, W=PΔV=nRΔTW = P\Delta V = nR\Delta T, ΔU=nCvΔT\Delta U = nC_v\Delta T. First law: Q=ΔU+WQ = \Delta U + W, so

nCpΔT=nCvΔT+nRΔT    CpCv=R.nC_p \Delta T = nC_v \Delta T + nR\Delta T \implies C_p - C_v = R.

Effective γ for Mixtures

For a mix of n1n_1 moles of gas with Cv1C_{v1} and n2n_2 moles with Cv2C_{v2}:

Cvmix=(n1Cv1+n2Cv2)/(n1+n2),C_v^\text{mix} = (n_1 C_{v1} + n_2 C_{v2})/(n_1 + n_2),

with similar formula for CpmixC_p^\text{mix}. Then γmix=Cpmix/Cvmix\gamma_\text{mix} = C_p^\text{mix}/C_v^\text{mix}. NEET often tests this with mono+diatomic mixtures.

Heat Engine and Pump Efficiencies

  • Refrigerator COP: β=QC/W=QC/(QHQC)\beta = Q_C/W = Q_C/(Q_H - Q_C). For Carnot, β=TC/(THTC)\beta = T_C/(T_H - T_C). Higher COP at lower ΔT\Delta T.
  • Heat pump COP (for heating): βHP=QH/W=1+βref\beta_\text{HP} = Q_H/W = 1 + \beta_\text{ref}.

Entropy (Conceptual)

The second law can be phrased: in any process, the total entropy of the universe increases (for irreversible) or stays the same (reversible).

For reversible heat exchange: dS=dQ/TdS = dQ/T.

For an ideal gas isothermal expansion: ΔS=nRln(Vf/Vi)\Delta S = nR\ln(V_f/V_i) — positive (irreversible if free expansion, reversible if isothermal but slow).

Phase Diagrams (Qualitative)

A typical P-T diagram for water shows three regions (solid, liquid, gas), separated by phase boundaries meeting at the triple point (273.16 K, 611 Pa). Above the critical point (647 K, 22 MPa), no distinction between liquid and gas.

Formula Sheet

QuantityFormula
Celsius–KelvinTK=TC+273.15T_K = T_C + 273.15
Linear expansionΔL=L0αΔT\Delta L = L_0 \alpha \Delta T
Volume expansionΔV=V0γΔT\Delta V = V_0 \gamma \Delta T (γ=3α\gamma = 3\alpha)
Heat (no phase change)Q=msΔTQ = ms\Delta T
Latent heatQ=mLQ = mL
Fourier's lawdQ/dt=KAΔT/LdQ/dt = KA\Delta T/L
Thermal resistanceR=L/(KA)R = L/(KA)
Stefan–BoltzmannP=ϵσAT4P = \epsilon\sigma A T^4
Wien's lawλmT=b\lambda_m T = b
Newton's coolingdT/dt=k(TT0)dT/dt = -k(T - T_0)
Ideal gasPV=nRTPV = nRT
Pressure (KT)P=13ρv2P = \tfrac{1}{3}\rho\overline{v^2}
Avg KE (monatomic)32kBT\tfrac{3}{2}k_BT
RMS speedvrms=3kBT/m=3RT/Mv_\text{rms} = \sqrt{3k_BT/m} = \sqrt{3RT/M}
Mean free pathλ=1/(2πd2n)\lambda = 1/(\sqrt 2\pi d^2 n)
MayerCpCv=RC_p - C_v = R
Isothermal workW=nRTln(V2/V1)W = nRT\ln(V_2/V_1)
AdiabaticPVγ=PV^\gamma = const
Adiabatic workW=(P1V1P2V2)/(γ1)W = (P_1V_1 - P_2V_2)/(\gamma - 1)
Carnot efficiencyη=1TC/TH\eta = 1 - T_C/T_H
Refrigerator COPβ=TC/(THTC)\beta = T_C/(T_H - T_C)

Sub-topics

6 pages

Practice quiz

Quiz
NEET Unit 8: Thermodynamics and Kinetic Theory — Quiz
15 questions · pick the best answer
Q1

The efficiency of a Carnot engine operating between 400 K and 300 K is:

Q2

Ratio of vrms:vavg:vpv_rms :v_avg :v_p for an ideal gas at temperature T is approximately:

Q3

For a diatomic gas at room temperature, γ=Cp/Cv= C_p/C_v equals:

Q4

Two rods of equal length and cross-section but conductivities K and 2K are joined in series. The effective conductivity is:

Q5

A black body radiates at temperature T. If T doubles, the radiated power becomes:

Q6

100 g of ice at 0 °C is mixed with 100 g of water at 80 °C. Final temperature is (Lf=80cal/g,swater=1(L_f = 80cal/g, s_water = 1 cal/g/°C):

Q7

Wien's displacement law: λmax_max T = constant. If T of a hot body doubles, λmax:_max:

Q8

Assertion: In an adiabatic process, no heat is exchanged with surroundings. Reason: The system is thermally insulated, so Q = 0.

Q9

For an ideal gas undergoing isothermal expansion, the change in internal energy is:

Q10

Mean free path λ of a gas molecule is inversely proportional to:

Q11

Work done in compressing 2 moles of ideal gas isothermally from V to V/2 at 300 K is:

Q12

If γ = 5/3 for a monatomic gas, the number of degrees of freedom is:

Q13

A body at 100 °C cools to 80 °C in 10 minutes in a room at 20 °C. Using Newton's cooling, time to cool from 80 °C to 60 °C (approximately) is:

Q14

Average translational KE of a gas molecule at temperature T is:

Q15

Assertion: A refrigerator cools its inside by extracting heat from inside and dumping it outside. Reason: This requires external work, in accordance with the second law of thermodynamics.