Look at figure 19-32 at the end of chapter 19. In the process of taking a gas fr

Question

Look at figure 19-32 at the end of chapter 19. In the process of taking a gas from state a to state c. along the curved path, 80 J of heat leave the system and 55 J of work are done on the system. We write Q_a rightarrow c = -80 J and W_a rightarrow c = -55 J. Determine the change in internal energy. U_a - U_c. When the gas is taken along the path cda, the work done on the gas W = 38 J. How much heat Q is added to the gas in the process cda? If P_a = 2.5P_d, how much work is done by the gas in the process abc? What is Q for path abc? If U_a - U_b = 10 J, what is Q for the process be?

Explanation / Answer

First Law of Thrermodynamics state
Heat gained by the system = Change in its internal energy + work done by the sytem
=> - 80 = Uc - Ua - 55
=> Ua - Uc = 80 - 55 J = 25.0 J.

b)
Q
= Ua - Uc + W
= 25 + 38 J
= 63.0 J.

c)
Work done for the path cda = 38 J (given)
=> Pd * (Vd - Vc) = 38 J
=> (Pa/2.44) * (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) * (2.5) = - 95.0 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)
= - 63 - 95 J
= - 158 J.

e)
Ua - Ub = 10 J and Uc - Ua = - 25 J
Adding, Uc - Ub = - 15 J
Q(bc)
= Uc - Ub + W (bc)
= - 15 + 0
= - 15 J
=> negative sgn indicates that heat leaves the system

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

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