A quasi-static cycle of a monoatomic ideal gas contains an isothermal process ( ab ), followed by an isochoric process ( bc ) and an adiabatic process ( ca ) as shown in the figure. The volumes of the gas are V_1 and V_2 at a and b , respectively. If the cycle has heat input Q_ in and output Q_ out , then the efficiency of the cycle is defined as = Q_ in - Q_ out Q_ in . The correct statement(s) is/are: [Given: 2 0.7]
C. If V_2/V_1 = 8, then the temperature of the gas at a is 4 times the temperature of the gas at c .
For the monoatomic ideal gas, = 5 3 . Process ab is isothermal, so T_a = T_b and V_a = V_1, V_b = V_2. Process bc is isochoric, so V_b = V_c = V_2. Process ca is adiabatic, so T_c V_c^ -1 = T_a V_a^ -1 . T_c V_2^ 2/3 = T_a V_1^ 2/3 T_c = T_a ( V_1 V_2 )^ 2/3 . If V_2 V_1 = 8, T_c = T_a ( 1 8 )^ 2/3 = T_a 4 T_a = 4 T_c. Thus, statement (C) is correct. Heat absorbed in isothermal process ab is Q_ ab = nRT_a ( V_2 V_1 ). For V_2 V_1 = 8, Q_ ab = nRT_a (8) = 3nRT_a 2 3 0.7 nRT_a = 2.1 nRT_a. Heat released in isochoric process bc is Q_ bc = nC_v(T_b - T_c) = n ( 3 2 R ) (T_a - T_a 4 ) = 9 8 nRT_a = 1.125 nRT_a. Since 1.125 nRT_a The efficiency of the cycle is = 1 - Q_ out Q_ in = 1 - Q_ bc Q_ ab . = 1 - n ( 3 2 R ) T_a (1 - ( V_1 V_2 )^ 2/3 ) nRT_a ( V_2 V_1 ) = 1 - 3 (1 - ( V_1 V_2 )^ 2/3 ) 2 ( V_2 V_1 ) . This expression depends only on the volume ratio V_2 V_1 and is independent of the temperature T_a. Thus, statement (B) is correct. For the isothermal process ab, P_a V_a = P_b V_b P_a V_1 = P_b V_2 P_a = 8 P_b. Thus, statement (D) is incorrect. Answer: If V_2/V_1 = 8, the heat released in the process bc is smaller than the heat absorbed in the process ab .; For a given value of V_2/V_1, does not depend on the temperature of the isothermal process.; If V_2/V_1 = 8, then the temperature of the gas at a is 4 times the temperature of the gas at c .
Related: Physics — Thermodynamics · All PYQ Banks