標題:
Physical Chemistry
發問:
1. A sample of 1.00 mol of an ideal gas at 1.00 atm and 298 K with Cpm = 7/2 R is put through the following cycle: (a) constant-volume heating to twice its initial temperature; (b) reversible, adiabatic expansion back to its initial temperature; (c) reversible isothermal compression back to 1.00 atm. Calculate q,... 顯示更多 1. A sample of 1.00 mol of an ideal gas at 1.00 atm and 298 K with Cpm = 7/2 R is put through the following cycle: (a) constant-volume heating to twice its initial temperature; (b) reversible, adiabatic expansion back to its initial temperature; (c) reversible isothermal compression back to 1.00 atm. Calculate q, w, deltaU and deltaH for each step ant for the overall cycle 2. An average human produces about 10 MJ of heat each day through metabolic activity. If a human body were an isolated system of mass 65kg with the heat capacity of water (Cpm = 75.3 JK-1mol-1), what temperature rise would the body experience? Human bodies are actually open systems, and the main mechanism of heat loss is through the evaporation of water (delta vap H = 44 kJmol-1). What mass of water should be evaporated each day to maintain constant temperature?
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1. 此循環的三個步驟如下圖所示： 圖片參考：http://imgcld.yimg.com/8/n/AF03901095/o/161009220148413872608460.jpg Cp = (7/2)R , Cv = Cp - R = (5/2)R , γ = Cp/Cv = 7/5 = 1.4 P1 = 1 atm T1 = 298 K V1 = RT1/P1 = 0.082x298/1 = 24.436 L V2 = V1 = 24.436 T2 = 298 x 2 = 596 K P2 = RT2/V2 = 2 atm 絕熱過程 P2V2^γ = P3V3^γ 借用 P3V3 = P1V1 = 24.436 P3V3^γ = P2V2^γ = 2x24.436^1.4 P3V3xV3^(γ-1) = 2x24.436^1.4 24.436xV3^(γ-1) = 2x24.436^1.4 V3 = (2x24.436^0.4)^(1/0.4) = 138.231 L 因此 P3 = 24.436/138.231 = 0.177 atm T3 = T1 = 298K (a) 1→2 恆容升溫 ?U = Cv (T2 - T1) = (5/2) x 8.314 x ( 596 - 298 ) = 6193.93 J w = ∫PdV = 0 系統不外界作功，外界也不對系統作功。 q = ?U + w = 6193.93 J ?H = ?U + ?(PV) = ?U + V?P = 6193.93 + 24.436x10^-3 (m^3) x 1(atm)x101325 (N/m2-atm) = 8669.91 J (b) 2→3 絕熱膨脹 q = 0 = ?U + w ?U = Cv (T3 - T2) = (5/2) x 8.314 x ( 298 - 596 ) = -6193.93 J w = -?U = 6193.93 J ?H = ?U + ?(PV) = ?U + (P3V3 - P2V2) = -6193.93 + ( 0.177x138.231 - 2x24.436)x101.325 = -8669.91 J (c) 3→1 恆溫壓縮 ?U = Cv ( T1 - T3 ) = 0 恆溫下，內能變化為零 w = RT ㏑(V1/V3) = 8.314 x 298 x ㏑(24.436/138.231) = -4293.31 J q = ?U + w = -4293.31 J ?H = ?U + (P1V1 - P3V3) = 0 循環總體變化 ?U = 0 w = 6193.93 - 4293.31 = 1900.62 J q = 6193.93 - 4293.31 = 1900.62 J ?H = 0 2. 65kg = 65000 g = 65000/18 = 3611 mol m x Cpm x ?T = Q ?T = 10x10^6/(3611x75.3) = 36.777 K/day 亦即若人體為絕熱系統，人體體溫將每天上升36.777度Ｋ 10x10^6 / 44000 = 227.273 mol = 4091 g H2O
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