heat must be supplied. Because 40=0, the first law takes the form @=w. This is a process that has 100%o conversion of heat into work atm The work exerted by the system is given by Work received. w Work=∫PdV here 2 and i denote the two states at the beginning and end of the process. The equation of state for an ideal gas is P= NRT/ with n the number of moles of the gas contained in the chamber. Using the equation of state, the expression for work can be written as Work during an isothermal expansion= nRTdv/V =NRTIn (A.1.1) For an isothermal process, PV=constant, so that P/P2=v2/V,. The work can be written in terms of the pressures at the beginning and end as Work during an isothermal expansion= NRTin (A.1.2) P The lowest pressure to which we can expand and still receive work from the system is atmospheric pressure. Below this, we would have to do work on the system to pull the piston out further There is thus a bound on the amount of work that can be obtained in the isothermal expansion; we of continuous power, in other words a device that would give power or propulsion as long as fuel was added to it. To do this, we need a series of processes where the system does not progres through a one-way transition from an initial state to a different final state, but rather cycles back to the initial state. What is looked for is in fact a thermodynamic cycle for the system We define several quantities for a cycle Qa is the heat absorbed by the system OR is the heat rejected by the system w is the net work done by the system. The cycle returns to its initial state, so the overall energy change, 40, is zero. The net work done by the system is related to the magnitudes of the heat absorbed and the heat rejected by W=Net work =2A-OR The thermal efficiency of the cycle is the ratio of the work done to the heat absorbed.(Efficiencies are often usefully portrayed as "What you get"versus"What you pay for. Here what we get is work and what we pay for is heat, or rather the fuel that generates the heat. In terms of the heat absorbed and rejected the thermal efficiency is Work done n= thermal efficiency Heat absorbed 4-QR=19k1A-2 P, T Q Work received, W Patm heat must be supplied. Because ∆U = 0, the first law takes the form Q=W. This is a process that has 100% conversion of heat into work. The work exerted by the system is given by Work = PdV 1 2 ∫ where 2 and 1 denote the two states at the beginning and end of the process. The equation of state for an ideal gas is P = NRT/V, with N the number of moles of the gas contained in the chamber. Using the equation of state, the expression for work can be written as Work during an isothermal expansion = NRT dV V/ 1 2 ∫ = NRTln V V 2 1       . (A.1.1) For an isothermal process, PV = constant, so that PP V V 12 21 / / = . The work can be written in terms of the pressures at the beginning and end as Work during an isothermal expansion = NRTln P P 1 2       . (A.1.2) The lowest pressure to which we can expand and still receive work from the system is atmospheric pressure. Below this, we would have to do work on the system to pull the piston out further. There is thus a bound on the amount of work that can be obtained in the isothermal expansion; we cannot continue indefinitely. For a power or propulsion system, however, we would like a source of continuous power, in other words a device that would give power or propulsion as long as fuel was added to it. To do this, we need a series of processes where the system does not progress through a one-way transition from an initial state to a different final state, but rather cycles back to the initial state. What is looked for is in fact a thermodynamic cycle for the system. We define several quantities for a cycle: QA is the heat absorbed by the system QR is the heat rejected by the system W is the net work done by the system. The cycle returns to its initial state, so the overall energy change, ∆U , is zero. The net work done by the system is related to the magnitudes of the heat absorbed and the heat rejected by W QQ = =− A R Net work . The thermal efficiency of the cycle is the ratio of the work done to the heat absorbed. (Efficiencies are often usefully portrayed as “What you get” versus “What you pay for”. Here what we get is work and what we pay for is heat, or rather the fuel that generates the heat.) In terms of the heat absorbed and rejected, the thermal efficiency is: η = thermal efficiency = Work done Heat absorbed = − = − Q Q Q Q Q A R A R A 1 . (A.1.3)