A) 3.0 x 10^1 J B) 68 J C) -84 J D) -114 J E) -15 J (III) In the process of taki

Question

A) 3.0 x 10^1 J B) 68 J C) -84 J D) -114 J E) -15 J

(III) In the process of taking a gas from state a to state c along the curved path shown in Fig. 19-32, 85 J of heat leaves the system and 55 J of work is done on the system. (a) Determine the change in internal energy, Eint,a - Eint,c. (b) When the gas is taken along the path cda, the work done by the gas is W = 38 J. How much heat Q is added to the gas in the process cda? (c) If Pa = 2.2d, how much work is done by the gas in the process abc? (d) What is Q for path abc? (e) If Eint,a - Eint,b = 15 J, what is Q for the process bc? Here is a summary of what is given: Q a c = -85 J W a c = -55 J W cda = 38 J Eint,a - Eint,b = 15 J Pa = 22d.

Explanation / Answer

a)
First Law of Thrermodynamics state
Heat gained by the system = change in its internal energy + work done by the sytem
=> - 85.0 = Uc - Ua - 55
=> Ua - Uc = 85.0 - 55.0 J = 30 J = 3.0*10^1 J

b)
Q
= Ua - Uc + W
= 30 + 38 J
= 68 J

c)
Work done for the path cda = 38 J (given)
=> Pd * (Vd - Vc) = 38J
=> (Pa/2.2) * (Vd - Vc) = 38
=> Pa * (Va - Vb) = 38... [because Vd - Vc = Va - Vb]
=> work done by the gas in the process abc
= Pa * (Vb - Va) = - (38) * (2.2) = -83.6 J = -84 J
Negative sign indicates that work is done on the gas in compressing its volume from Va to Vb (=Vc).

d)
Q (for path abc)
= Uc - Ua + W (for path abc)
= - 30 - 84 J
= - 114 J.

e)
Ua - Ub = 15 J and Uc - Ua = - 30 J
Adding, Uc - Ub = - 15 J
Q(bc)
= Uc - Ub + W (bc)
= - 15 + 0
= - 15 J
=> negative sgn indicates that heat leaves the system.

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