Question

2.00 mol of neon is expanded isothermically at room temperature from 2.00 atm to 1.00 atm and is then heated from room temperature to 125 C at a constant pressure of 1.00 atm.

a) What is the change in entropy for the overall process?

b) What is the change in entropy for the process with the same initial and final states as in part (a), but in which the gas is first heated at constant pressure and then expanded isothermically?

c) Draw the two processes on two different graphs choosing the proper thermodynamic state variables to represent them.

(Please show how you got your answers. One of them should be 23.55)

Answer 1

at Constant pressure AS = cp lu ! Calculate change in Entropy dor Isothermal Exporria of Ideal gas As == nR ln P22 where n=2 moles R=0.0821 Formula to calculate change in Entropy Pi = 2 atm ; P2=latm p = 20.70 (Neon) Ti Tz = 125 +273 k Change in Entropy for over all process T,= 25+273 k pRoom Torp) As = -nRlu P + cp lu I =-2X0.0821 lu I + 20.79 du 39 As = 0.1138 +5.92 OS: 6.0338 For Both procon ss will be same because it a state function 5 For Isothermal proces For Isobaric Exponsion zatmt - P I atm.