University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 19 - The First Law of Thermodynamics - Problems - Exercises - Page 643: 19.44

Answer

See explanation.

Work Step by Step

(a) To know the heat transfer, Q, we need to first identify the $\Delta U $ and $W$ . W is the area under the p versus V curve. $W = \frac{1}{2} (1.0 ×10^5 Pa + 3.5 ×10^5 Pa)(0.0060 m^3 − 0.0020 m^3 ) + \frac{1}{2}(1.0 ×10^5 Pa + 3.5 ×10^5 Pa)(0.0100 m^3 − 0.0060 m^3 )$ $W = 1800 J $ To find $\Delta U$ we need to identify the changes in temperature of the gasses. $\Delta T = T_c - T_a = \frac{p\Delta V}{nR}$ $\Delta T = \frac{(1.0 ×10^5 Pa) (0.01 m^3 - 0.002m^3)}{(1/3)(8.314 J/mol.K)}$ $\Delta T = 289 K$ $\Delta U = nC_V \Delta T$ $\Delta U = (1/3)(12.47 J/mol ⋅K)(289 K)$ $\Delta U = 1201 J$ $Q = ΔU +W$ $Q = 1201 J + 1800 J $ $Q = 3001 J$ The heat is transferred into the gas. (b) Step a to c is isobaric which means the pressure is kept constant. We can straight away use the equation $Q = nC_p \Delta T$ $Q = (1/3)(20.79 J/mol ⋅K)(289 K)$ $Q = 2002 J$ The heat is transferred into the gas because Q > 0 (c) Q is larger in part (a) than part (b). This means Q in closed loop like stated in part (a) has more heat energy in the system and did more work. It is expected to be more efficient.
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