The Figure shows a cycle undergone by 1.00 mol of an ideal monatomic gas. For 1

Question

The Figure shows a cycle undergone by 1.00 mol of an ideal monatomic gas. For 1 rightarrow 2, what are heat Q, the change in internal energy Delta E_int, and the work done W? For 2 rightarrow 3, what are Q, Delta E_int and (f) W? For 3 rightarrow 1, what are Q, Delta E_jnt, and (i) W? For the full cycle, what are Q, Delta E_int, and The initial pressure at point 1 is 1.00 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

(a) from 1 to 2 volume is constant so the process is isochoric thus work done is zero in this process now change in internal energy is heat dQ = dU = nCvdT (here n = 1 mol Cv = 3/2R = 12.465J/kg and dT = T2 - T1 = 300K ) thus dQ = 1*12.465*300 = 3739.5J

(b) change in internal energy is same as heat

(c) work done is zero in this process from 2 to 3 the process is adiabatic therefore gamma (r) = 5/3 and there is no exchange of heat between system and surrounding therefore

(d) dQ = 0 in this case

(e) internal enerty dU = - dW = nR(T3 -T2)/(1-r) = 1*8.31*(455-600)(1-5/3) = -1807.4 J

(f)work is W = nR(T3-T2) = 1807.4 J for 3 to 1 the process is isobaric therefore P is constant

(g) we know that the net change in internal energy is zero for a cyclic process therefore.

heat dQ = nCpdT (here n = 1 mol Cp = 5/2R = 20.785 J/kg and dT = T3 - T1 = 155K )

thus dQ = 1*20.785*155 = -3221.675 J

h)

i)

j) 517.825 J

k) Zero

l) 517.825 J

m) V2 = nRT1/P1 = 0.0246 m^3

n) P2 = nRT2/V2 = 2.028*10^5 Pa = 2 atm

o) V3 = nRT3/P3 = 0.0373 m^3

p) p3 = p1 = 1.013*10^5 Pa = 1 atm

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