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1st Law of Thermodynamics The change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work performed by the system on its surroundings. (Version of conservation of energy applied to thermodynamical systems. Thermodynamic Processes: Isothermal Using Boyle’s Law – To keep the temperature constant both the pressure and volume change to compensate. (Volume goes up, pressure goes down) Change in internal energy is a function of ΔT. If ΔT = 0, then ΔU = 0 Then Q = W W = nRTln(Vf/Vi) ln is the natural log function Thermodynamic Processes: Isobaric No change in pressure: ΔP = 0 Heat is added to the gas which increases the Internal Energy (U). Work is done by the gas as it changes in volume. On a PV diagram, the isobaric process is a horizontal line called an isobar. ∆U = Q - W can be used since the WORK is POSITIVE in this case ∆U = Q – W where W = PΔV Thermodynamic Processes: Isovolumetric No change in volume: ΔV = 0 On a PV diagram, an isovolumetric process is a vertical line. No area under the curve, and no work done. ΔU = Q Thermodynamic Processes: Adiabatic System exchanges no heat with its surroundings Q=0 PV diagram is a steep hyperbola. ΔU = - W Example Problem 13.17 Two moles of an ideal gas expands isothermically at 27°C to three times its initial volume. (a) Find the work done by the gas (b) The heat flow into the system Example Problem 13.22 A 1-kg block of aluminum is heated at atmospheric pressure such that its temperature increases from 22°C to 40°C. Find: (a) The work done by the aluminum (a) The heat added to the aluminum (b) The change in internal energy of the aluminum Example Problem 13.54 A gas follows the path 123 on the PV diagram below and 418 J of heat flows into the system. Also, 167 J of work is done. Pressure, P 2 1 3 4 Volume, V Find the internal energy of the system: Example Problem 13.54 A gas follows the path 123 on the PV diagram below and 418 J of heat flows into the system. Also, 167 J of work is done. Pressure, P 2 1 3 4 Volume, V How much heat flows into the system if the process follows the path 143? The work done by the gas along this path is 63 J: Example Problem 13.54 A gas follows the path 123 on the PV diagram below and 418 J of heat flows into the system. Also, 167 J of work is done. Pressure, P 2 1 3 4 Volume, V What net work would be done on or by the system if the system followed the path 12341 ?