Consider the expansion of 60.0 moles of a monatomic ideal gas along processes 1
ID: 2157944 • Letter: C
Question
Consider the expansion of 60.0 moles of a monatomic ideal gas along processes 1 and 2 in Figure 18-27. In process 1 the gas is heated at constant volume from an initial pressure of Pi = 123 kPa to a final pressure of Pf = 246 kPa. In process 2 the gas expands at constant pressure from an initial volume of Vi = 1.20 m3 to a final volume of Vf = 3.60 m3.
(a) How much heat is added to the gas during these two processes?
kJ (process 1)
kJ (process 2)
(b) How much work does the gas do during this expansion?
kJ
(c) What is the change in the internal energy of the gas?
kJ
Explanation / Answer
Ti = PiVi/nR = 123*10^3*1.2/(60*8.314) = 296 K
a) For process 1:
W = 0 since constant volume
Eint = nCvdT = 60*2.5*8.314*(296*246/123 - 296) = 369000 J = 369 kJ
Q = W+Eint = 369 kJ
For process 2:
W = 246*(3.6-1.2) = 590.4 kJ
Eint = 60*2.5*8.314*[(296*246/123)*3.6/1.2 - (296*246/123)] = 1476566 J = 1476.6 kJ
Q = W+Eint = 2067 kJ
b) W = 590.4 kJ
c) Eint = 369+1476.6 = 1845.6 kJ