General Physics III Practice Exam I Solutions Fall 2007

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2. A heat engine uses 1.00 mol of a monatomic gas and follows the cycle shown in the diagram below. The temperature of state 1 is 300 K. The energy transferred by heat to the gas during the process 1 to 2 is 3750 J. (a) Calculate the volume of the gas for state 1, and the pressures, volumes and temperatures for the gas at states 2, 3 and 4. Show your work and list the values in the table below. State P (kPa) V (m 3 ) T (K) 1 300 8.31 x10 -3 300 2 601 8.31 x10 -3 601 3 601 1.66 x10 -2 1202 4 300 1.66 x10 -2 600 (a) 12 points (1pt for V 1 ,V 2 , V 3 ,V 4 , P 2 , P 3 ) (2 pts. for T 2 , T 3 , T 4 ) (a) The first law of thermodynamics is Q =Δ Eth + WS . For the isochoric process 1 → 2, W S 1 → 2 = 0 J. Thus, Q = 1 2 3750 J = Δ E = th nC Δ → V T 3750 J 3750 J 3750 J ⇒Δ T = = = = 301 K 3 3 nCV ( 1.0 mol)( 2 R) ( 1.0 mol)( 2)( 8.31 J/mol K) ⇒T − T = 300.8 K ⇒ T = 300.8 K + 300 K = 601 K To find volumes V 2 and V 1 , V 2 1 2 ( 1.0 mol )( 8.31 J/mol K)( 300 K) −3 3 nRT = V = = = 8.31× 10 m 3.0 10 Pa 1 2 1 5 p1 × The pressure p 2 can be obtained from the isochoric condition as follows: p p T ⎛ = ⇒ p = p = ( ) T T T ⎜ × = × 300 K ⎟ ⎝ ⎠ 2 1 2 2 1 2 1 1 601 K ⎞ 3.00 10 5 Pa 6.01 10 5 Pa With the above values of p 2 , V 2 and T 2 , we can now obtain p 3 , V 3 and T 3 . We have The value of p 4 = p 3 , and V 4 = V 1 ; and V = 2V = 1.662 × 10 m p = p = 6.01× 10 Pa −2 3 5 3 2 3 2 T T V = ⇒ T = T = 2T = 1202 K V V V 3 2 3 3 2 2 3 2 2 T T V = ⇒ T = T = 2T = 600 K V V V 4 1 4 4 1 1 4 1 1
(b) Determine the work done by the gas system W S , the heat Q, and the change in internal energy for each of the four processes in the cycle. Show your work and list your answers in the table below. Process W S (J) Q (J) ΔE int (J) 1 → 2 0 3750 3750 2 → 3 4990 12,480 7490 3 → 4 0 –7500 –7500 4 → 1 –2490 –6230 –3740 (b) (17 Points) (1 pt. for W 12 , W 34 , Q 34, ΔE int12. ΔE int34 ), (2 pts. for W 23 , W 41 , Q 13 , Q 41 , ΔE int23. ΔE int41 ) For the isochoric process 1 → 2, W S 1 → 2 = 0 J. Thus, ΔE th = Q 12 = 3750 J For the isobaric process 2 → 3, 5 5 ( )( )( ) ( )( )( )( ) Q = nC Δ T = 1.0 mol R T − T = 1.0 mol 8.31 J/mol K 601 K = 12,480 J 2→3 P 2 3 2 2 S 2→3 3 3 2 We are now able to obtain p 4 , V 4 and T 4 . We have 5 −3 3 ( ) ( )( ) W = p V − V = 6.01× 10 Pa 8.31× 10 m = 4990 J Δ E = Q − W = 12,480 J − 4990 J = 7490 J th 2→3 S 2→3 V = V = 1.662 × 10 m p = p = 3.00 × 10 Pa −2 3 5 4 3 4 1 5 T4 T3 p ⎛ 4 3.00 × 10 Pa ⎞ = ⇒ T4 = T3 = ⎜ 5 ⎟( 1202 K) = 600 K p4 p3 p3 ⎝ 6.01× 10 Pa ⎠ For isochoric process 3 → 4, 3 3 ( )( )( ) ( )( )( )( ) Q = nC Δ T = 1.0 mol R T − T = 1.0 mol 8.31 J/mol K − 602 = −7500 J 3→4 V 2 4 3 2 W = 0 J ⇒ Δ E = Q − W = −7500 J S 3→4 th 3→4 S 3→4 For isobaric process 4 → 1, Q = nC Δ T = 1.0 mol 5 8.31 J/mol K 300 K − 600 K = −6230 J 4→1 P 2 S 4→1 4 1 4 ( ) ( )( ) 5 −3 3 −2 3 ( ) ( ) ( ) W = p V − V = 3.00 × 10 Pa × 8.31× 10 m − 1.662 × 10 m = −2490 J th 4→1 S 4→1 ( ) Δ E = Q − W = −6230 J − − 2490 J = −3740 J (c) Calculate the efficiency of this heat engine and compare it to the efficiency of a Carnot heat engine operating between the same maximum and minimum temperatures as this cycle. (5 points) The thermal efficiency of this heat engine is η W W 2500 J Q Q + Q 3750 J + 12,480 J out out = = = = = H 1→2 2→3 0.154 15.4% The efficiency of a Carnot heat engine operating between the same maximum and minimum temperatures would be (4 points) TC 300K e = 1 − C 1 1 0.25 0.75 75% T = − 1200K = − = = H
Page 1: General Physics III Practice Exam I
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General Physics III Practice Exam I Solutions Fall 2007

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