Question

Consider a Carnot cycle in which the working substance is 0.10 mol of perfect gas molecules, the temperature of the hot source is 373 K, that of the cold sink is 273 K; the initial volume of gas is 1.00 dm', which doubles over the course of the first isothermal stage. For the reversible adiabatic stages it may be assumed that VT3/2 = constant. a) calculate the volume of the gas VB and Vc); b) calculate the volume of the gas after stage 3 (VD) by considering the reversible adiabatic compression from the starting point; c) for each of the four stages of the cycle, calculate the heat transferred to or from the gas; d) Explain why the work done is equal to the difference between the heat extracted from the hot source and that deposited in the cold sink; e) calculate the work done over the cycle and hence the efficiency n: f) confirm that your answer agrees with the efficiency from n = 1-T/T).

Answer 1

A The Carnot cycle. > A~B isothermal expansion. Bol adiabatic expansion . 1" D o isothermal compression. ro> DNA adiabatic compression. Stage I. temperature Say the in this case is Ts The process is isothermal, so du zo. HERO A dwa -pdv a We - RRT - NRT, en la & & & .W & NRT, en v a feat input. du Stage I The process is adiabatic, hence da 20. W = AU NCAT & new (T2-T) T2 Stage in The process is isothermal so again, Wot - nRT₂ in to & Heat output - W . NRT₂ im Vo ve Stag Il. The process is adiabatic & temperature changes from T2 to To W e werd nou (T. -T2 ) TTT2. Hence overall work done wa We + W2 + Wg + Wa Nongtime = -ART, in VB & Now (T2-T ) - NRT₂ mVD & nc (T2-T1) V. VA o 8, +82 Now 8, is taken as heat output, hence heat released by the system. So if we take the magnitude of heart released as &2, then the heat released shouldere

active sign. Hence w e & d, where where 8, 4 d2 both are positive quantity & or W = Heat absorbed the magnitudes of the heat a - heat released = (heat to extracted from) - (he heat deposited to Hot source the cold sink Ans.(d). Now for adiabatic reversible process, given, Tra constant for Bc. ** TRV Tevens for DA - T zu 2 Vc given VA a 1 dm3 a 1it. - alt. Voo double the initial volume Jaib Tiz 3T3K Now x 373 373 272 => Vc a 2x² 273 2.706 U. 2.71 lt Now, Vo ve . * 2.71 4.1.355 Hence VB. alt. San Ve 2.71 4. Ans: (a) na

Voo 1.355 lt Ans (6) . for A-SB, Heat transf-8, RRT, as in v A ered to gas Ord mol x 8.314 Tintik x 373K * in. > 214.95 J. This was heat input for Bye & Dan A heat input & or Since they son adiabatic processes. are zero input & output heat DNA ah For CAD, Vc from gas? Heat released 8, . Net n e nRT₂ in D 1.355 I O.LX 8.314 x 273 x in. 2.71 2 -157 31 J This is heat output hence ave sign occurs if we take magnitude of the then 157.31 J. Ans: (C) The work done over the cycle. - - MR (Tz en vo je Tim VA) a 8,- da a 214.05-157.34 T 57.64 J. Efficiency Iwl an. NR (T2 en Up y - To MVA) Sinput NRT in VB VA - NRT, in ve nk in (T-12 nk im VB r. VB . Ve] VA vol T T (1)
Where T_h =Temperature of Hot source, T_c=Temperature of the cold sink