ch20-p035

# ch20-p035 - 35. (a) The pressure at 2 is p2 = 3.00p1, as...

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(e) The process 4 1 is adiabatic, so 44 11 p Vp V γγ = and 41 1.30 14 1 0.165, (4.00) pV γ §· == = ¨¸ ©¹ where we have used V 4 = 4.00 V 1 . (f) The efficiency of the cycle is ε = W / Q 12 , where W is the total work done by the gas during the cycle and Q 12 is the energy added as heat during the 1 2 portion of the cycle, the only portion in which energy is added as heat. The work done during the portion of the cycle from 2 to 3 is W 23 = ³ pdV . Substitute 22 p pV V = to obtain () 3 2 23 2 2 2 3 . 1 V V Wp VV d V V V −− ³ 35. (a) The pressure at 2 is p 2 = 3.00 p 1 , as given in the problem statement. The volume is V 2 = V 1 = nRT 1 / p 1 . The temperature is 2 21 1 3.00 3.00 3.00. T TT nR nR T = ¡ = (b) The process 2 3 is adiabatic, so 33 TV = . Using the result from part (a), V 3 = 4.00 V 1 , V 2 = V 1 and γ =1.30, we obtain 1 0.30 2 12 3 1 3.00 3.00 1.98 /3.00 4.00 V V = = . (c) The process 4 1 is adiabatic, so = . Since V 4 = 4.00 V 1 , we have 1 0.30 1 0.660. 4.00 = ¨¸¨¸ (d) The process 2 3 is adiabatic, so p V = or 32 3 2 p VV p = . Substituting V 3 = 4.00 V 1 , V 2 = V 1 , p 2 = 3.00 p 1 and =1.30, we obtain 3 1.30 1 3.00

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## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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ch20-p035 - 35. (a) The pressure at 2 is p2 = 3.00p1, as...

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