Sample Thermodynamic Problems (test prep) p. 503 50) Gas within a chamber passes through the cycle shown in the figure to the right. How much energy is transferred as heat during the process CA? The energy added as heat during the process AB is 20.0 Joules. The process BC is adiabatic. The net work done during the complete cycle is 15.0 Joules. ANS: Since the process is a complete cycle (beginning and ending in the same thermodynamic state) the change in the internal energy is zero and the heat absorbed by the gas is equal to the work done by the gas: Q = W. In terms of the contributions of the individual parts of the cycle QAB + QBC + QCA = W and QCA = W – QAB – QBC = +15.0 J – 20.0 J – 0 = –5.0 J. This means 5.0 J of energy leaves the gas in the form of heat. P. 533 46) One mole of an ideal diatomic gas goes from a to c along the diagonal path shown in the figure to the left. The scale of the vertical axis is set by pab = 5.0 kPa and pc = 2.0 kPa. The scale of the horizontal axis is set by Vbc = 4.0 m3 and Va = 2.0 m3. During the transition, A) What is the change in internal energy? B) How much energy is added as heat? ANS: Two formulas (other than the first law of thermodynamics) will be of use to us. It is straightforward to show, from Eq. 19-11, that for any process that is depicted as a straight line on the pV diagram — the  pi  p f Wstraight    2 work is   V  which includes, as special cases, W = pV for constant-pressure processes and W = 0 for constantvolume processes. Further, Eq. 19-44 with Eq. 19-51 gives  f   f  Eint  n   RT    pV 2 2 where we have used the ideal gas law in the last step. We emphasize that, in order to obtain work and energy in Joules, pressure should be in Pascals (N / m2) and volume should be in cubic meters. The degrees of freedom for a diatomic gas is f = 5. (a) The internal energy change is 5 5  pcVc  paVa     2.0 103 Pa  4.0 m3   5.0 103 Pa  2.0 m3   2 2 3  5.0 10 J. Eint c  Eint a  (b) The work done during the process represented by the diagonal path is  p  pc  3 3 Wdiag   a  Vc  Va  =  3.5 10 Pa  2.0 m   2  which yields Wdiag = 7.0×103 J. Consequently, the first law of thermodynamics gives Qdiag  Eint  Wdiag  (5.0 103  7.0 103 ) J  2.0 103 J. p. 558 41) The figure to the right represents a Carnot engine that works between temperatures, T1 = 400 K and T2 = 150 K. The engine drives a Carnot refrigerator that works between temperatures T3 = 325 K and T4 = 225 K. What is the ratio of Q3/Q1? ANS: The efficiency of the engine is defined by  = W/Q1 and is shown in the text to be  T1  T2 W T1  T2 .   T1 Q1 T1 The coefficient of performance of the refrigerator is defined by K = Q4/W and is shown in the text to be T4 Q4 T4 . K   T3  T4 W T3  T4 Now Q4 = Q3 – W, so (Q3 – W)/W = T4/(T3 – T4). The work done by the engine is used to drive the refrigerator, so W is the same for the two. Solve the engine equation for W and substitute the resulting expression into the refrigerator equation. The engine equation yields W = (T1 – T2)Q1/T1 and the substitution yields Q Q3T1 T4 = 3 1 =  1. T3  T4 W Q1 T1  T2  Solving for Q3/Q1, we obtain   T  T   T   T  T  1  T2 T1  Q3  T4   1  1 2    3   1 2   . Q1  T3  T4   T1   T3  T4   T1  1  T4 T3  With T1 = 400 K, T2 = 150 K, T3 = 325 K, and T4 = 225 K, the ratio becomes Q3/Q1=2.03. 42A) During each cycle, a Carnot engine absorbs 750 Joules as heat from a high-temperature reservoir at 360 K, with the low-temperature reservoir at 280 K. How much work is done per cycle? ANS: Eq. 20-13 gives the Carnot efficiency as 1 – TL /TH . This gives 0.222 in this case. Using this value with Eq. 20-11 leads to W = (0.222)(750 J) = 167 J. 42B) The Carnot engine in 42A) is made to work in reverse to function as a Carnot refrigerator between those same two reservoirs. During each cycle, how much work is required to remove 1200 Joules of heat from the low-temperature reservoir? ANS: Now, Eq. 20-16 gives KC = 3.5. Then, Eq. 20-14 yields |W| = 1200/3.5 = 343 J.