Please help solve part (d) and (e) for the following problem (I found parts (a),

Question

Please help solve part (d) and (e) for the following problem (I found parts (a), (b), and (c)): "In the process of taking a gas from state a to state c along the curved path shown in (Fig. 15-29 BELOW), 88 J of heat leave the system and 57 J of work are done on the system. Here is a summary of what is given: Qac = -88 J , Ub - Ua = -10 J , Wac = -57 J , Pa = 3.5 ? Pd

(a) Determine the change in internal energy, Uc - Ua. ANSWER: -31J

(b) When the gas is taken along the path abc, the work done by the gas is W = -125 J. How much heat Q is added to the gas in the process abc? ANSWER: -156J

(c) If Ub- Ua= -10 J, what is Q for the process bc?   ANSWER: -10J

(d) If Pa = 3.5Pd, how much work is done by the gas in the process cda?

(e) What is Q for path cda?

Explanation / Answer

d) In the process cda the work done by the gas is
W = P dV
W = Pd x (Vd - Vc)
For the process da no work is done because it is a constant volume process dV =0
Pd = Pa / 3.5
W = (Pa / 3.5 ) x Vd - Vc
But Vd = Va and Vc = Vb
W = Pa x (Va - Vb) / 3.5
W = - Wabc / 3.5
Wabc is given as -125J
W = 125 / 3.5 = 35.71 J
e) Q for the path cda
Q = change in internal energy + work done
Q =( Ua - Uc ) + Wcda
Ua - Uc = - (Uc - Ua) = 31J (from part a of the question)
Wcda  = 35.71 J
Q = 31 J + 35.71 J = 66.71 J

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