P21_033 - 1 ´ = − µ p 1 V 1 γ − 1 ¶µ 1 − 1 4 γ...

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33. (a) The pressure at 2 is p 2 =3 p 1 , as given in the problem statement. The volume is V 2 = V 1 = nRT 1 /p 1 . The temperature is T 2 = p 2 V 2 nR = 3 p 1 V 1 nR =3 T 1 . The process 4 1 is adiabatic, so p 4 V γ 4 = p 1 V γ 1 and p 4 = µ V 1 V 4 γ p 1 = p 1 4 γ , since V 4 =4 V 1 . The temperature at 4 is T 4 = p 4 V 4 nR = µ p 1 4 γ ¶µ 4 nRT 1 p 1 ¶µ 1 nR = T 1 4 γ 1 . The process 2 3 is adiabatic, so p 2 V γ 2 = p 3 V γ 3 and p 3 =( V 2 /V 3 ) γ p 2 . Substitute V 3 =4 V 1 , V 2 = V 1 ,and p 2 =3 p 1 to obtain p 3 = 3 p 1 4 γ . The temperature is T 3 = p 3 V 3 nR = µ 1 nR ¶µ 3 p 1 4 γ ¶µ 4 nRT 1 p 1 = 3 T 1 4 γ 1 , where V 3 = V 4 =4 V 1 =4 nRT/p 1 is used. (b) The efficiency of the cycle is ε = W/Q 12 ,whe re 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 = R p d V . Substitute p = p 2 V γ 2 /V γ to obtain W 23 = p 2 V γ 2 Z V 3 V 2 V γ dV = µ p 2 V γ 2 γ 1 ³ V 1 γ 2 V 1 γ 3 ´ . Substitute V 2 = V 1 , V 3 =4 V 1 ,and p 3 =3 p 1 to obtain W 23 = µ 3 p 1 V 1 1 γ ¶µ 1 1 4 γ 1 = µ 3 nRT 1 γ 1 ¶µ 1 1 4 γ 1 . Similarly, the work done during the portion of the cycle from 4 to 1 is W 41 = µ p 1 V γ 1 γ 1 ³ V 1 γ 4 V 1
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Unformatted text preview: 1 ´ = − µ p 1 V 1 γ − 1 ¶µ 1 − 1 4 γ − 1 ¶ = − µ nRT 1 γ − 1 ¶µ 1 − 1 4 γ − 1 ¶ . No work is done during the 1 → 2 and 3 → 4 portions, so the total work done by the gas during the cycle is W = W 23 + W 41 = µ 2 nRT 1 γ − 1 ¶µ 1 − 1 4 γ − 1 ¶ . The energy added as heat is Q 12 = nC V ( T 2 − T 1 ) = nC V (3 T 1 − T 1 ) = 2 nC V T 1 , where C V is the molar speciFc heat at constant volume. Now γ = C p /C V = ( C V + R ) /C V = 1 + ( R/C V ), so C V = R/ ( γ − 1). Here C p is the molar speciFc heat at constant pressure, which for an ideal gas is C p = C V + R . Thus, Q 12 = 2 nRT 1 / ( γ − 1). The efficiency is ε = 2 nRT 1 γ − 1 µ 1 − 1 4 γ − 1 ¶ γ − 1 2 nRT 1 = 1 − 1 4 γ − 1 ....
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