Answer to Question #190720 in Molecular Physics | Thermodynamics for Ahmed

Question #190720

Suppose you are given a one mole of ideal gas, in an initial state p=6 atm, v=3L, and in a clock wise direction, you maneuvered the system reverisbly in a path given by (v-6)2

+(p-6)2 =16 in the pv plane

Deduce the amount of work done by the gas as a result of this process

Find out the change if the internal energy of the gas

Calculate the minimum and maximum temperature attained by the gas during the cycle



1
Expert's answer
2021-05-10T09:23:09-0400

p=6 atmV=3 Lp = 6\ atm\\ V = 3\ L

path=(V6)2+(p6)2=16path = (V-6)^2 +(p-6)^2 =16


maximum value of V when p = 6 atm

(V6)²+(66)²=16(V6)²=16V6=4V=10 L(V-6)² + (6-6)² = 16\\ (V-6)² = 16\\ V-6 = 4\\ V= 10\ L


maximum value of p when V = 3 L

(36)²+(p6)²=16(3)²+(p6)²=169+(p6)²=16(p6)²=7p6=7p=6+7=8.65 atm(3-6)²+ (p -6)² = 16\\ (-3)² +(p-6)² =16\\ 9+ (p-6)²= 16\\ (p-6)² = 7\\ p-6 =\sqrt{7}\\ p= 6 +\sqrt7= 8.65 \ atm


Work done = 12(VfVi)(pfpi)\dfrac12(V_f-V_i)(p_f-p_i)


=12(103)(8.656)=9.275 J=\dfrac12(10-3)(8.65-6) = 9.275\ J



U=QW.D=09.275J∆U = Q-W.D = 0-9.275J

U=9.275 J\therefore ∆U = -9.275\ J

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