Question

Five moles of carbon dioxide (CO2), initially 3 atm and 300 K, is trapped inside a piston-cylinder assembly. It is then allowed to expand against atmospheric pressure adiabatically. The constant- pressure heat capacity of CO2 is given by the following equation: p = 5.4574 (1.045 × 10-3)(T/K)-(1.157 × 105)(T/K)-2 (a) If we assume that the expansion is infinitely slow and quasi-static, calculate the final temperature and total volume of the carbon dioxide gas when it reaches 1 atm. Also calculate the work done in the process. (b) The other extreme is to assume sudden expansion. Repeat your calculations in Part (a) in this scenario

Answer 1

five n_co_2 = B moles P_1 = 3 atm T_1 = 300k By ideal gaslaw, V_1 = nRT_1/P_1 = 5 times 8.314 times 300/3 times 101325 = 0.041 m^3 there is no heat exchange with the surroundings integral dq = 0 = integral^T_T_1 ncpoT = 0 c_p = 45.37 + 8688 times 10^-3T - 0.619 times 10^3/T^2 (since R = 8.314 J/mol.k implies = 5 integral^T_T1(45.37 + 86.688 times 10^-3T- (9.62 times 10^5/T^2)dt implies 8[226.85(T - 300) + 0.0435/2(T^2 -(300)^2) + 48.1 times 10^3[1/T - 1/100] = 0 implies 226.85(T - 300) + 0.0435/2(T^2 -(-300)^2) implies [T - 300][226.85 times 300T + 0.02175(T + 300) - 4810000] = 0 \r\n 68055T + 6.525 + 0.02175T - 4810000 = 0 T_2 = 70.68K P_2 = 1 atm v_2 = nRT_2 = 5 times 8.314 times 70.68/101325 V_2 = 0.029 m^3 r_co_2 = 1.28 w_adiabatic = P_1V_1 - P_2V_2/ partial differentiate - 1 = (3 times 0.041 - 1 times 0.029)101325/1.28 - 1 = (0.3357) times 101325 = 34.016 KJ w = 34.016KJ/mol b. For an free expansion process, \r\n