A 2.35 mole sample of an ideal gas, for which C v, m= 3R/2, initially at 27.0 de
ID: 594805 • Letter: A
Question
A 2.35 mole sample of an ideal gas, for which C v, m= 3R/2, initially at 27.0 degree C and 1.75 times 106 Pa undergoes a two-stage transformation. For each of the stages described in the following list, calculate the final pressure, as well as Q, w, Delta U and Delta H. Also calculate Q, w, Delta U and Delta H for the complete process. The gas is expanded isothermally and reversibly until the volume triples. Beginning at the end of the first stage, the temperature is raised to 105.0 degree C at constant volume.Explanation / Answer
No. of moles = 2.35 , Cv,m = 3R / 2 , T1 = 27 C , P1 = 1.75 x 10 ^6 pa
A) Isothermal reversible
An isothermal process is a change of a system, in which the temperature remains constant: T = 0 SO T2 = 27C
PV = Constant ==> P1V1 = P2V2
As given V2 = 3V1 so
P2 = P1V1 / 3V1 = P1 / 3 = 5.83 x 10^5 Ans
U = 0 ,H = 0 So W = -PdV = -6.44 x 10 ^3 J, q = 6.44 x 10 ^3 Ans
b) T3 = 105 C,Constant volume so P/T = Constant
P2/T2 = P3/T3
P3 = P2 * T3 / T2 = 7.35 x 10 ^3 Ans
dV =0 so W=0,U = 2.29 x 10^3J , H = CvdT = 3.81 x 10^3 J
As w = 0 So q = U = 2.29 x 10^3J
C) Overall q = qa + qb = 8.73 x 10 ^3 J
Work w = Wa + wb = -6.44 x 10 ^3 J
U = Ua + Ub = 2.29 x 10^3J
H = Ha + Hb = 3.81 x 10^3 J
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