A 10 cm diameter cylinder contains argon gas at 10 atm pressure and a temperatur

Question

A 10 cm diameter cylinder contains argon gas at 10 atm pressure and a temperature of 45C . A piston can slide in and out of the cylinder. The cylinder's initial length is 23cm . 2400J of heat are transferred to the gas, causing the gas to expand at constant pressure. Part A What is the final temperature of the cylinder? Part B What is the final length of the cylinder?

Explanation / Answer

a) The molar heat capacity at constant volume of an ideal monatomic gas is cv_m = Cv/n = (3/2)·R Divide this by the specific heat capacity (Cv/m) and you have the molar mass of the gas: M = m/n = Cv_m/Cv = (3/2)·8.3145J/molK / 625J/kgK = 0.01995 kg/mol ˜ 20g/mol Compare this to known molar masses He 4g/mol, Ne 20g/mol , N 14g/mol actually 28g/mol cause N is diatomic, Ar 40g/mol So the answer is Neon b) Heat transferred is an constant pressure process is equal to the change of enthalpy. Change of enthalpy of an ideal gas is given by ?H = n·cp_m·?T = n·cp_m·(T_final - T_initial) for any ideal gas cp,m - cv;m = R so for a monatomic ideal gas cp,m = (5/2)·R Q = n·(5/2)·R·(T_final - T_initial) => T_final = T_initial + Q/(n·(5/2)·R) The number of moles can be found from ideal gas law V_initial = (p/4)·D²·L = (p/4)· (0.1m)²·0.21m = 0.00165m³ n = p ·V_initial / (R · T_initial ) = 10·101325Pa · 0.00165m³ / (8.3145J/molK · (55+273)K ) = 0.613mol => T_final = 328K + 2500J/(0.613mol · (5/2) · 8.3145J/molK) = 524K = 251°C c) Because number of moles in the cylinder and pressure dont change: V/T = n·R/p = constant => V_initial / T_initial = V_final/T_final (p/4)·D²·L_initial / T_initial = (p/4)·D²··L_final / T_final => L_final = L_initial · (T_final / T_initial) = 21cm · (524K / 328K) = 33.5cm

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