This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Okay, Blessing, let's break down these physics problems. Part 15b(i): Coefficient of linear expansion Step 1: Determine the change in temperature and fractional change in length. The temperature change is T = 100^ - 0^ = 100^. The percentage increase in length is 0.05\%, so the fractional increase in length is ( L)/(L_0) = (0.05)/(100) = 0.0005. Step 2: Calculate the coefficient of linear expansion (). The formula for linear expansion is L = L_0 T. Rearranging for : = ( L)/(L_0 T) = (0.0005)/(100^) = 0.000005 /^ = 5.0 × 10^-6 /^ The coefficient of linear expansion is: 5.0 × 10^-6 /^ Part 15b(ii): Percentage increase of the volume of the bar Step 1: Relate the coefficient of volume expansion () to the coefficient of linear expansion (). For isotropic materials, the coefficient of volume expansion is approximately three times the coefficient of linear expansion: = 3 = 3 × (5.0 × 10^-6 /^) = 1.5 × 10^-5 /^ Step 2: Calculate the percentage increase in volume. The fractional increase in volume is ( V)/(V_0) = T. ( V)/(V_0) = (1.5 × 10^-5 /^) × (100^) ( V)/(V_0) = 0.0015 To express this as a percentage: Percentage increase = 0.0015 × 100\% = 0.15\% The percentage increase of the volume of the bar is: 0.15\% Part 16a(i): Quantity that remains constant for thermodynamic processes A) Isobaric process: Pressure* (P) B) Isochoric process: Volume* (V) C) Isothermal process: Temperature* (T) D) Adiabatic process: Heat exchange* (Q=0) Part 16a(ii): Show that PV^ = constant for a monatomic ideal gas undergoing an adiabatic process Step 1: State the First Law of Thermodynamics for an adiabatic process. For an adiabatic process, there is no heat exchange, so dQ = 0. The First Law of Thermodynamics states dQ = dU + dW. Therefore, dU + dW = 0. Step 2: Express dU and dW for an ideal gas. For an ideal gas, the change in internal energy is dU = nC_v dT, where n is the number of moles and C_v is the molar specific heat at constant volume. The work done by the gas is dW = P dV. Substituting these into the equation from Step 1: nC_v dT + P dV = 0 (1) Step 3: Use the ideal gas law to relate dT to dP and dV. The ideal gas law is PV = nRT. Differentiating both sides: P dV + V dP = nR dT Rearranging for dT: dT = (P dV + V dP)/(nR) (2) Step 4: Substitute dT into equation (1) and simplify. Substitute (2) into (1): nC_v ((P dV + V dP)/(nR)) + P dV = 0 (C_v)/(R) (P dV + V dP) + P dV = 0 We know that R = C_p - C_v, where C_p is the molar specific heat at constant pressure. So, (C_v)/(C_p - C_v) (P dV + V dP) + P dV = 0. Divide by C_v: (1)/(C_p)C_v - 1 (P dV + V dP) + (P dV)/(C_v) = 0 Let = (C_p)/(C_v) (the adiabatic index). (1)/( - 1) (P dV + V dP) + P dV = 0 Multiply by ( - 1): P dV + V dP + ( - 1) P dV = 0 V dP + P dV + P dV - P dV = 0 V dP + P dV = 0 Step 5: Integrate the differential equation. Divide the equation by PV: (dP)/(P) + (dV)/(V) = 0 Integrate both sides: (dP)/(P) + (dV)/(V) = 0 P + V = constant P + V^ = constant (PV^) = constant Exponentiating both sides: PV^ = e^constant Since e^constant is also a constant, we have: PV^ = constant Part 16b(i): Efficiency for a Carnot cycle A) As a function of the heat that flows to and from the reservoirs: = 1 - (Q_C)/(Q_H) where Q_C is the heat rejected to the cold reservoir and Q_H is the heat absorbed from the hot reservoir. B) As a function of the temperatures of the two reservoirs: = 1 - (T_C)/(T_H)