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.
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Here are the solutions to the problems: 10. Determine the value of the index of expansion n. Step 1: Write down the given values and the polytropic process equation. The process follows the law P_1 V_1^n = P_2 V_2^n. Given: P_1 = 480 kPa V_1 = 4.4 m^3 P_2 = 156.88 kPa V_2 = 10.4 m^3 Step 2: Rearrange the equation to solve for n. (P_1)/(P_2) = ((V_2)/(V_1))^n Taking the natural logarithm of both sides: ((P_1)/(P_2)) = n ((V_2)/(V_1)) n = ((P_1)/(P_2))((V_2)/(V_1)) Step 3: Substitute the given values and calculate n. n = (480 kPa156.88 kPa)(10.4 m^34.4 m^3) n = ((3.060938))/((2.363636)) n = (1.1189)/(0.8603) n ≈ 1.30 The value of the index of expansion is 1.30. 11. Calculate the value of the characteristic gas constant R. Step 1: Convert temperature to Kelvin and pressure to Pascals. Given: P = 120 kPa = 120 × 10^3 Pa T = 20^ C = 20 + 273.15 = 293.15 K = 1.68 kg/m^3 Step 2: Use the ideal gas law in terms of density to find R. The ideal gas law is P = R T. Rearrange to solve for R: R = (P)/( T) Step 3: Substitute the values and calculate R. R = 120 × 10^3 Pa(1.68 kg/m^3)(293.15 K) R = (120000)/(492.492) R ≈ 243.66 J/(kg · K) Converting to kJ/(kg · K): R ≈ 0.24366 kJ/(kg · K) The characteristic gas constant is 0.244 kJ/(kg · K). 12. Determine the specific heat capacities. Given: m = 1.2 kg T_1 = 15^ C = 15 + 273.15 = 288.15 K T_2 = 210^ C = 210 + 273.15 = 483.15 K Q = 160 kJ (at constant pressure) R = 0.242 kJ/(kg · K) 12.1 Specific heat capacity of the gas at constant pressure (c_p). Step 1: Calculate the change in temperature ( T). T = T_2 - T_1 = 483.15 K - 288.15 K = 195 K Step 2: Use the formula for heat transfer at constant pressure to find c_p. Q = m c_p T Rearrange to solve for c_p: c_p = (Q)/(m T) Step 3: Substitute the values and calculate c_p. c_p = 160 kJ(1.2 kg)(195 K) c_p = (160)/(234) c_p ≈ 0.684 kJ/(kg · K) The specific heat capacity at constant pressure is 0.684 kJ/(kg · K). 12.2 Specific heat capacity of the gas at constant volume (c_v). Step 1: Use Mayer's relation, which connects c_p, c_v, and R. c_p - c_v = R Rearrange to solve for c_v: c_v = c_p - R Step 2: Substitute the calculated c_p from 12.1 and the given R. c_v = 0.68376 kJ/(kg · K) - 0.242 kJ/(kg · K) c_v = 0.44176 kJ/(kg · K) The specific heat capacity at constant volume is 0.442 kJ/(kg · K). That's 2 down. 3 left today — send the next one.