PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/138MFpB2WKxnY4QckyPoa97KHwtiEeM2s/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true of state, latent heat; Thermal equilibrium, Zeroth law of thermodynamics; Heat, work and internal energy. First law of thermodynamics; Carnot engine and its efficiency HEAT When a hot body is kept in contact with a cold body, there is a transfer of energy from hot body to cold body. The energy transferred is called heat. ZEROTH LAW OF THERMODYNAMICS : If a body A is separately in thermal equilibrium with body B and body C then B and C are also in the thermal equilibrium. “Two bodies which are in thermal equilibrium are said to have equal temperatures”. Thermal Expansion When the temperature of a body increases, its size increases. Coefficient of linear expansion is given by CHAPTER COVERS : Thermal Expansion Coefficient of Apparent Expansion of a Liquid Specific Heat Molar Specific Heat Latent Heat Internal Energy First Law of α = ΔL LΔT Thermodynamic Different Lθ = L0 (1 + αθ) Coefficient of superficial expansion is given by Thermodynamic Processes Work Done by β = ΔA AΔT Aθ = A0 (1 + βθ) Coefficient of cubical expansion is given by ΔV System in Different Processes Indicator Diagram Slope of Isothermal and Adiabatic γ = VΔT or Vθ = V0 (1 + γ θ) Curves Carnot Engine and m = m (1 + γθ) ρθ ρ0 ⇒ ρθ ≈ ρ0(1 – γθ) Heat Pump Efficiency of Carnot An Isotropic body expands equally in all directions and we can obtain the following relations γ = 3α, β = 2α or Cycle and Carnot Theorem Applications If αx, αy, αz are coefficient of linear expansion along x-axis, y-axis and z-axis, (i.e., for anisotropic body) then γ = αx + αy + αz. For water γ is negative between 0°C and 4°C Density of water is maximum at 4°C. Therefore water at the bottom of lake in winter is warmer than that at the surface. Two rods of length L1 and L2 are kept side by side. If with increase in temperature, the difference in their lengths does not change i.e. L2′ − L1′ = L2 − L1 where L2′ = L1(1+ α1Δt ) and L2′ = L1(1+ α2 Δt ) then L α = L α or L1 = α2 . 1 1 2 2 L1 − L2 α2 − α1 L1, α1 L2, α2 A vessel is completely filled with a liquid αv = coefficient of linear expansion of vessel γl = coefficient of cubical expansion of liquid With increase in temperature, the volume of liquid flown out is given by ΔV = V0[1 + (γl – γv)Δθ] = V0[1 + (Vl – 3αv)Δθ] ⇒ ΔV = V0[1 + γaΔθ], where γa = coefficient of apparent expansion = γl – γv = γl – 3αv Variation of moment of inertia with temperature I' = I(1 + βΔθ) Liquid of volume V0(γl) Glass container (αv) of volume V0 Also, ΔI = βΔθ = − Δω  (When Angular momentum L = Iω = Constant) I ω As an annular disc is heated, all the dimensions (including cavity) increase. Thus, we have r1′ = r1 (1 + αΔθ) r2′ = r2 (1 + αΔθ) Annular disc In other words, you can say that the thermal expansion is photographic i.e. as when a photograph is enlarged, all the dimensions of the photograph increase. Coefficient of volume expansion of an ideal gas at constant pressure is given by γ = 1 T A meter scale callibrated at T1°C is used for measurement at T2°C. Let α be the coefficient of linear expansion for scale, and gives a reading L, then error in the reading is ΔL = αL(T2 – T1). CALORIMETRY Specific heat capacity : c = ΔQ mΔT (cal/g/°C) Molar heat capacity : C = ΔQ nΔT (cal/mol/°C) Latent heat of fusion Lf = m of vaporisation Lv = m Water Specific heat C = 1 cal/gm/°C = 4.2 J/gm/°C = 4200 J/kg/°C Lf = 80 cal/gm = 336 J/gm Lv = 540 cal/gm = 2268 J/gm Application To convert m mass of ice at 0°C into steam at 100°C, amount of heat required is 80 m + 100 m + 540 m = 720 m cal (a) Two objects having masses m1, m2, specific heat capacities c1, c2 and temperatures t1 and t2 are mixed. If there is no change in state during mixing, then resulting temperature of mixture, t = m1c1t1 + m2c2t2 mix m1c1 m2c2 (b) Specific heat of the mixutre, c = m1c1 + m2c2 mix m1 + m2 w gm of water at T°C is mixed with m gm of ice at 0°C. (a) (b) w = 80m T w < 80m T ⇒ Whole of ice melts. Final temperature = 0°C. ⇒ Whole of ice will not melt. Final temperature = 0°C Amount of ice melted, m′ = wT 80 (c) w > 80m T Amount of ice left = m – m′ ⇒ Whole of ice melts. Final temperature = wT − 80m > 0°C w + m KINETIC THEORY OF GASES Assumptions for ideal gas are Kinetic theory of gases is applicable for large number of molecules. Intermolecular forces between two molecules is negligible. The force due to gravity on the molecules is neglected. The separation between the molecules is much larger as compared to their size. The molecules are perfectly elastic and all collisions between the molecules and a wall are considered to be perfectly elastic. All gases at high temperature and low pressure behave like an ideal gas. Pressure Exerted by the Gas The pressure of the gas is due to continuous bombardment of the gas molecules against the walls of the container. According to kinetic theory, the pressure exerted by an ideal gas is given by M = Mass of the enclosed gas V = Volume of the container v 2 = Mean square speed of molecules Or P = 1 Nm v 2 N = Number of molecule 3 V m = Mass of the molecule v 2 = Mean square speed of molecules Types of speed defined for a gas : (i) Root mean square speed, vrms = = ρ = (ii) v Avg = = = v1 + v 2 + v3 + + vn n (iii) vmp = Most probable speed is defined the speed corresponding to which there are maximum number of molecules. vmp = = Order of magnitude : vrms > vavg > vmp ρ = Density of gas Mw = Molecular weight R = Gas constant P = Pressure of gas Vrms : Vav : Vmp = : : Kinetic Interpretation of Temperature Translation Kinetic Energy = 3 nRT 2 THERMODYNAMICS Internal Energy : Every bulk system consists of a large number of molecules. Internal energy is simply the sum of the kinetic energies and potential energies of these molecules. It is a macroscopic variable. For gases ΔU = f nRΔT = f (P V − PV ) 2 2 2 2 1 1 Gas Degrees of freedom (f) ΔU = f nRΔT 2 C = ΔU V nΔT CP=CV + R γ = CP CV Monoatomic 3 (Translational) 3 nRΔT 2 3 R 2 5 R 2 5 3 Diatomic or Linear Polyatomic 3(Trans) + 2(Rot) 5 nRΔT 2 5 R 2 7 R 2 7 5 Non-Linear Poly atomic 3 (Trans) + 3 (Rot) 3nRΔT 3 R 4 R 4 3 Meanfree Path (λ) The meanfree path is the average distance covered by a molecule between two Successive collision. λ = 1 n = No. of molecules per unit volume It is inversely proportional to density of gas. In closed container it temperature is increased λ remain same. First Law of Thermodynamics : It is law of conservation of energy. Let Q heat is supplied to gas. It is used in two ways. Increasing internal energy (i.e., increasing temperature). Work done by the gas during expansion (W) ⇒ Q = ΔU + W Thermodynamic Process Melting process : (Change of state, solid to liquid) Q = ΔU + W mLf = ΔU + 0 [W = 0 as volume remains nearly constant] Boiling process : (Change of state, liquid to gas) mLv = ΔU + P[V2 – V1] V2 = volume of vapours V1 = volume of liquid When 1 g of water vapourises isobarically at atmospheric pressure. ΔU = 2091 J, P = 1.01 × 105 Pa, V1 = 1 cm3, V2 = 1671 m3. Isochoric process : dV = 0 ⇒ W = 0 [dV = change in volume] Q = nCVΔT = ΔU ⇒ CV = ΔU nΔT Isobaric process : P = constant, dW = PdV ⇒ W = PΔV = nRΔT Q = nCPΔT = ΔU + W nCPΔT = nCVΔT + nRΔT ⇒ CP = CV + R ΔU or 1 = W = Q γ − 1 γ Fraction of total heat converted to internal energy = ΔU = 1 W Fraction of total heat converted to work is, Q Q γ = γ − 1 γ Isothermal process : PV = K ⇒ ΔT = 0 ⇒ ΔU = 0, C = ∞ To calculate the amount of work done by the gas in an isothermal process. as PV = nRT (Constant) So P = nRT V Work done W = Q = nRT log V2 = 2.303 nRT log = V2 e 10 1 1 Adiabatic process : PV γ = K [Equation of adiabatic process] As Q = 0, nCΔT = 0 or C = 0 Also, 0 = nCV ΔT + W [by first law of thermodynamics] Now, Polytropic Process PVn = Constant W = nRΔT 1− n C = CV R 1− n Indicator Diagram: P-V graph of a process is called indicator diagram. Area under P-V graph represents the work done in a process. P Isochoric Process : Slope = P = nR T V V W = 0 Isobaric : V V2 V1 Slope = V = nR T P V O T1 T2 T W = P(V2 – V1) = PΔV = nRΔT Isothermal process : P1 1 T = constant Slope = dP = − P 2 2 O V1 V V2 dV V Adiabatic process : = – γP P Applications P-V graph for different gases for adiabatic expansion P 3 2 γ = 4/3 1 γ = 7/5 γ = 5/3 V → monoatomic → diatomic → polyatomic P.V. graph for isothermal & adiabatic expansion & compression for a given gas P (a) Isothermal Adiabatic (b) V Expansion of a gas under different processes P 1 1 → Isobaric 2 2 → Isothermal 4 3 → Adiabatic 3 4 → Isochoric V |W1| > |W2| > |W3| Compression P 2 1 CARNOT ENGINE Carnot Cycle : 1 → Isobaric 3 4 2 → Isothermal → Adiabatic → Isochoric V |W3| > |W2| > |W1| In a carnot engine the working substance (an ideal gas) draws some heat from the source per cycle (say Q1) performs some work W per cycle and rejects heat Q2 to the sink per cycle. W P1, V1 V 1 → 2 Isothermal Expansion ΔU = 0 W = Q = nRT ln V2 (positive) 1 1 V 1 2 → 3 Adiabatic Expansion Q = 0 W = −ΔU = nRΔT 2 1− r 3 → 4 Isothermal Compression ΔU = 0 W = Q = nRT l ⎛ V4 ⎞ (negative) 3 2 n⎜ V ⎟ ⎝ 3 ⎠ 4 → 1 Adiabatic Compression Q = 0 W = −ΔU = nRΔT 4 1− r Heat supplied = Q1 Heat rejected = Q2 ⇒ Q1 – Q2 = W η = Wtotal × 100 = Q1 − Q2 × 100 Qsupplied Q1 ⎛ Q2 ⎞ ⎜1 − Q ⎟ × 100 ⎝ 1 ⎠ ⎛ T2 ⎞ = ⎜1 − T ⎟ × 100 (for ideal engine) ⎝ 1 ⎠ : Q : W = T : T : T − T or Q1 = Q2 = W 1 2 1 2 1 2 T1 T2 T1 − T2 Carnot Theorem : The efficiency of carnot engine is maximum (<100%) for given temperatures T1 and T2. Any other engine working between temperature range T1 & T2 cannot have efficiency more than the carnot engine working between the same temperature range. Heat Pump : In a heat pump W work is done on the working substance per cycle, Q2 heat is absorbed by the substance from lower temperature T2 per cycle and Q1 heat is supplied to higher temperature T1(T1 > T2) per cycle. β = Heat Supplied = Q1 = Q1 Wtotal W Q1 − Q2 β = T1 T1 − T2 (for ideal pump) Relation between η and β, η = 1 β + 1 and β = 1 − 1 η Refrigerator : In a refrigerator, W work is done on the working substance, Q2 heat is absorbed from lower temperature T2 and Q1 heat is rejected to higher temperature T1. (T1 > T2). Coefficient of performance β = heat rejected = Wtotal ⇒ β = T2 T1 − T2 Q2 Q1 − Q2 ❑ ❑ ❑