Question

The Sackur-Tetrode Equation gives the entropy of a sample of n moles of monatomic ideal gas as a function of its internal energy U and volume V S(U, V) = 5/2 n R + n R In (V/n N_A(4piM U/3nN^2_Ah^2)^3/2) In the equation, R is the gas constant, M is the molar mass, N_4 is Avogadro's number, and h is Plank's constant. The equation can be derived using S = k ln W and directly computing W, the number of microstates consistent with the given volume and internal energy. The development of the equation was delayed by the fact that when the molecules are assumed to follow the laws of classical physics the number of microstates is infinite. The breakthrough required came from quantum mechanics. By assuming the molecules had a minimum linear size, delta x, and a minimum momentum range, delta p, that were related by delta c delta p approximately h, the number of microstates was made finite. This assumption is the Heisenberg uncertainty principle and is the reason Planck 's constant appears in the Sackure Tetrode equation. Show that the Sackure Tetrode equation can be written more simply as S(U, V) = n beta + R in (V/n) + 3/2 n R in (U/n). Where Beta is a constant that depends only on the molar mass and fundamental constants. Evaluate B for argon gas. (Answer: 83.07 J/mol. K) Invert the Sackure Tetrode equation to obtain an expression for the internal energy of a gas as a function of entropy and volume, U(S, V). Evaluate the internal energy of 0.040 moles of argon gas occupying a volume of 0.0010 ms and having an entropy of 5.8 J/K. (Answer: 67 J) The first law of thermodynamics can be written in differential form as dU = Tds - PdV. Knowing the internal energy as a function of entropy and volume one can use this to find the temperature and pressure as function of entropy and volume as well. The temperature is given by T = partial U/partial S)_v. Use this to evaluate the temperature of 0.040 moles of argon gas occupying a volume of 0.0010 m^3 and having an entropy of 5.8 J/K. (Answer: 130 K) The pressure then is given by P = - (partial U/partial V)_s. Use this to evaluate the pressure of 0.040 moles of argon gas occupying a volume of 0.0010 m^3 and having an entropy of 5.8 J/K. (Answer: 45 kPa)

Answer 1

S(U,N) = 5/2nR + nRln (V/nN_A(4piMU/3nN^_Ah^2)^3/2) = 5/2nR + nRln(V/n) + nRln(1/N_A(4piMU/3nN^2_Ah^2)^3/2) = 5/2nR + nRln(V/n) + 3/2nRln(4piMU/3nN^2_Ah^2N^2/3_A) = 5/2nR + nRln(V/n) + 3/2nRln(U/n) + 3/2 nRln(4piM/3h^2N8/3_A) = [5/2R + 3/2Rln(4piM/3h^2N^8/3_A)]h + nRln(V/n) + 3/2nRln(U/n) where beta = 5/2R + 3/2Rln (4piM/3h^2N^8/3_A) putting values R = 8.314 J/mol: k h = 6.626 times 10^-34 JS N_A = 6.023 times 10^23 beta = 83.07J/mol.K \r\n 2) S = nbeta + nRln(V/n) + 3/2nRln(U/n) S - n beta/nR = ln(V/n) + (U/n)^3/2 S - n beta/nR = ln(v/n(U/n)^3/2) exp(S - n beta/nR) = V/n(U/n)^3/2 U = n[n/v exp(S - n beta/nR)]^2/3 U = 0.04[0.04/0.001 exp(5.8 - 0.04 times 83.07/0.04 times 8.314)]^2/3 U = 67.1J 3) T = (2U/2S)_v T = 2U/3nR = 2(67.1)/3(0.04)(8.314) = 134.5k \r\n 4) P = -(2U/2V)_s using U = 3/2 nRT P = nRT/V = (0.04)(8.314)(134.3)/0.001 P = 44729.32 pa P tildeequal 45 kpa