He figure shows a cycle undergone by 1.38 mol of an ideal monatomic gas. For 1 t

Question

He figure shows a cycle undergone by 1.38 mol of an ideal monatomic gas. For 1 to 2, what are (a) heat Q, (b) the change in internal energy delta E_int and (c) the work done W? For 2 to 3, what are (d)Q, (e) AE_int, and (f)W? For 3 to 1, what are (g)Q, (h) delta E_int, and (i)W? For the full cycle, what are (k)delta E_int, and (l)W? The initial pressure at point 1 is 1.76 atm (where 1 atm = 1.013 times 10^-5 Pa). What are the (m) volume and (n) pressure at point 2 and the (o) volume and (p) pressure at point 3?

Explanation / Answer

1 ---> 2

Workdone by gas = 0 because the volume is constant.

change in internal energy = nC_v delta_T = 1.38*3*8.314*300/2 = 5162.994 J(for monoatomic C_v = 3R/2

hence from first law of thermodynamics Q = W+delta U = 5162.994 J

a) Q = 5162.994J

b) delta E = 5162.994 J

c) W = 0

2 ----> 3

for monoatomic gas   = 1.66

hence work done W = (PfVf-PiVi)/(1- ) = nR(Ti-Tf)/( -1) = 1.38*8.314*(600-455)/(1.66-1) = 2520.654 J

change in internal energy = 1.5nR deltaT = 1.5*1.38*8.314*(-600+455) = -2495.447 J

hence Q = -2495.447 + 2520.654 =  25.207 J

d) Q = 25.207 J

e) delta E = -2495.447 J

f) W = 2520.654 J

3 ----> 1

work done W = nR(Tf-Ti) = 1.38*8.314*(300-455) = -1778.365 J

change in internal energy = 1.5nR deltaT = 1.5*1.38*8.314*(300-455) = -2667.547 J

Hence Q = -1778.365 -2667.547 = -4445.912 J

g) Q = -4445.912 J

h) delta E = -2667.547 J

i) W = -1778.365 J

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.