Cuối kỳ Vật lý 2( Tiếng Anh )

FINAL PHY2 1 2 n = 0 1 moles V = 2 5

FINAL PHY2 1/ 2/ n = 0.1 moles V = 2.5 𝑑𝑚3 = 2.5 𝑥 10−3 𝑚3 3V = 7.5 𝑑𝑚3 = 7.5 𝑥 10−3 𝑚3 P1 = 105 𝑃𝑎 𝑅= 25 𝐽/(𝑚𝑜𝑙𝑒 𝐾) 𝛾 = 1.5 a) T1 = 𝑃𝑉 𝑛𝑅 = 105 𝑥 (2.5 𝑥 10−3 ) 0.1 𝑥 8.314 = 300.7 𝐾 P1V1 = P2V2 (Chỗ giống isothermal(Đẳng nhiệt)) Nhiệt độ lúc chưa thay đổi mà có áp suất) V(after the expansion) = V + 3V = 10 𝑑𝑚3 = 10 𝑥 10−3 𝑚3 P1V1 = P2V2 105 𝑥 (2.5 𝑥 10−3 ) = 𝑃2 𝑥 (10 𝑥 10−3 )  P2 = 25000 Pa b) Adibatic process : 𝑇1 𝑉2𝛾−1 = 𝑇2 𝑉3𝛾−1 V2 = (10 𝑥 10−3 ) 𝑚3 V3 = ( 2.5 𝑥 10−3 ) 𝑚3 (Because the piston back to its initial position)  300.7 𝑥 (10 𝑥 10−3 )1.5−1 = 𝑇2 𝑥 (2.5 𝑥 10−3 )1.5−1  T2 = 601.4 K 3/ T = 273 K R = 8.314 P = 1.0 𝑥 10−2 𝑎𝑡𝑚 = 1013.25 Pa d = 1.24 𝑥 10−5 𝑔/𝑐𝑚3 = (1.24 x 10^-5) x 10^6 : 10^3 = 0.0124 kg/m^3 Pascal với kg/m^3 a) vrms = √ 3𝑅𝑇 𝑀 =√ 3𝑃 𝑑 = √ 𝑥 0.082 𝑥 273 𝑀 =√ 𝑥 1013.25 0.0124 = 495 m/s b) vrms = √ 3𝑅𝑇 𝑀 −→ 𝑀 = 0.028 kg/mol = 28 g/mol c) M = 28g/mol  N2 4/ n = mole P(A) = 𝑃0 = 105 Pa P(B) = P0 = 105 𝑃𝑎 P(C) = 3P0 = x 105 Pa V(A) = V(B) = V0 = 𝑚3 a) From A  B (Isochoric = V constant) V(A) = V(B) = 𝑚3 From B  C (Isothermal = T constant) P(B) x V(B) = P(C) x V(C)  105 x = x 105 x V(C)  V(C) = 1/3 𝑚3 b) Adibatic process from C to A  P(C) x 𝑉 (𝐶 )𝛾 = 𝑃(𝐴) 𝑥 𝑉(𝐴)𝛾 105  (3 x 105 ) 𝑥 ( )𝛾 = 1  ( )𝛾 =  𝛾 = log = 1.631 𝑥 1𝛾 5/ At very low temperatures, the molar specific heat Cv of many solids is approximately Cv = 𝐴𝑇 , where A depends of the particular substance For aluminum, A = 3.15 𝑥 10−5 𝐽 𝑚𝑜𝑙 −1 𝐾 −4 Find the entropy change for 4.0 mol of aluminum when its temperature is raised from 5.0 K to 10.0 K The formula for an entropy change: dS = 𝒅𝑸 Equation: dQ = NCv(T + dT – T)  dQ = 𝑁𝐶𝑣 (𝑇)𝑑𝑇 (2) (1)(2)  dS = 𝑁𝐶𝑣(𝑇)𝑑𝑇 𝑇 Integrate both sides  ∆𝑆 = 𝑁 ∫ We have : Cv(T) = 𝐴𝑇 𝐶𝑣(𝑇)𝑑𝑇 𝑇 𝑻 (J) (1)  ∆𝑆 = 𝑁 ∫ 𝐴𝑇 𝑑𝑇 𝑇 = 𝑁𝐴 ∫ 𝑇 𝑑𝑇 = 𝑁𝐴 = x (3.15 x 10^-5) x 103 −53 ∆𝑇 3 = NA 𝑇23 −𝑇13 = 0.03675 J Other question 5: 𝑇2 𝑑𝑄 The formula of entropy change : ∆S = ∫𝑇1 𝑇 dQ = mc𝑑𝑇 𝑇2 𝑚(𝑎+𝑏𝑇)𝑑𝑇 ∆S = ∫𝑇1 𝑇 𝑇2 𝑚𝑎 = ∫𝑇1 𝑇 + 𝑚𝑏 = 𝑚𝑎𝑙𝑛(𝑇) + 𝑚𝑏𝑇 = x 770 x ln dS = 𝑑𝑄 𝑇 (J/K) (1) dQ = mCvdT (2) (1)(2)  dS = 𝑚𝑐𝑑𝑇 𝑇 With c = a+bT Thus : 𝑇2 𝑚(𝑎+𝑏𝑇)𝑑𝑇 Integrate both side : ∆𝑆 = ∫𝑇1 𝑇 𝑇2 = ∫𝑇1 (𝑚𝑎𝑇 −1 + 𝑚𝑏)𝑑𝑇 = maln(T) + mbT = maln(600/300) + mb(600-300) =3x ... P(C) = 3P0 = x 105 Pa V(A) = V(B) = V0 =