Consider the expansion of 60.0 moles of a monatomic ideal gas along processes 1
ID: 1521639 • Letter: C
Question
Consider the expansion of 60.0 moles of a monatomic ideal gas along processes 1 and 2 in the figure below. In process 1 the gas is heated at constant volume from an initial pressure of Pi = 95 kPa to a final pressure of Pf = 190 kPa. In process 2 the gas expands at constant pressure from an initial volume of Vi = 0.90 m3 to a final volume of Vf = 2.70 m3.
(a) How much heat is added to the gas during these two processes?
(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
a)
Process 1 :
Ti = Pi Vi / (nR) = 95000 x 0.90 / (60 x 8.314) = 171.4 K
Tf = Pf Vf / (nR) = 190000 x 2.70 / (60 x 8.314) = 1028.4 K
Pi/Ti = P1 /T1
95 /171.4 = 190 /T1
T1 = 342.8 k
W1 = 0 Since Volume remain constant
U1 = (3/2) n R (T1 - Ti) = (1.5) (60) (8.314) (342.8 - 171.4) = 128.25 kJ
Q1 = W1 + U1 = 128.25 kJ
Process 2 :
W2 = Pf (Vf - Vi) = 190,000 (2.70 - 0.90) = 342000 = 342 kJ
U2 = (3/2) n R (Tf - T1) = (1.5) (60) (8.314) (1028.4 - 342.8 ) = 513.01 kJ
Q2 = W2 + U2 = 342 + 513.01 = 855.01 kJ
b)
Work done = W1 + W2 = 0 + 342 kJ = 342 kJ
c)
Total internal energy change = U = U1 + U2 = 128.25 + 513.01 = 641.3 kJ