Question

6. Consider a system at T1, P1,V1 which is composed of one mole of a classical ideal gas for which PV=RT and U=A+BT. Let the system change in four steps. The first step is a very slow compression to P2,V2 at a constant temperature of T1. The second step is a quick compression to T3, P3, V3. The third step is a very slow expansion to P4, V4 at a constant temperature T3. The fourth step is a very quick expansion back to T1, P1, V1. The second and fourth steps are so quick that there is no heat exchange. Calculate the relations between the total work, heat of the first step and the heat of the third step, expressing the results in terms of T1 and T3.

Answer 1

For process (3-4) & (2-3) adiabatic so we can write T_4/T_1 = (V_1/V_4) ^r-1, T_3/T_ = (V_2/V_3) ^r-1 but T3 = T4 & T1 = T2 (V1/V4) ^r-1 = (V2/V3) ^r-1 Therefore V1/V2 = V4/V3 we can write for net work done Wnet = R(T3-T1) ln (v1/v2) Also efficiency is given by eta = Wnet/Q1 = R(T3-T1) ln(v1/v2) /R(T3) ln(v4/v3) v1/v2 = v4/v3 Therefore eta = T3-T1/T3 Therefore eta = 1-T1/T3\r\n The system composed of 1 Mole ideal gas for which PV = RT & V = A + BT System changes in four steps process two very slow isothermal process & two very fast adiabatic process As we know that internal energy of system is always depends on temperature Here internal is given by U = A + BT therefore dv/dT = B The change in internal energy w.r.t temperature is constant which indicates all process considered to be reversible process. The given data indicates analysis of cannot cycle the cycle is shown on P-V & T-S \process(1-2) Reversible isothermal compression In this process heat is rejected to heat sink with constant temperature Process(2-3) Reversible adiabatic process In this process by consuming eternal work, a gas is compresses by PV^p = C, here temperature of gas