First Law of Thermodynamics Conservation of Energy for Thermal Systems

Joule Equivalent of Heat James Joule showed that mechanical energy could be converted to heat and arrived at the conclusion that heat was another form of energy. He showed that 1 calorie of heat was equivalent to J of work. 1 cal = J

Energy Mechanical Energy: KE, PE, E Work is done by energy transfer. Heat is another form of energy. Need to expand the conservation of energy principle to accommodate thermal systems.

1 st Law of Thermodynamics Consider an example system of a piston and cylinder with an enclosed dilute gas characterized by P,V,T & n.

1 st Law of Thermodynamics What happens to the gas if the piston is moved inwards?

1 st Law of Thermodynamics If the container is insulated the temperature will rise, the atoms move faster and the pressure rises. Is there more internal energy in the gas?

1 st Law of Thermodynamics External agent did work in pushing the piston inward. W = Fd =(PA)  x W =P  V xx

1 st Law of Thermodynamics Work done on the gas equals the change in the gases internal energy, W =  U xx

1 st Law of TD Let’s change the situation: Keep the piston fixed at its original location. Place the cylinder on a hot plate. What happens to gas?

Heat flows into the gas. Atoms move faster, internal energy increases. Q = heat in Joules  U = change in internal energy in Joules. Q =  U

1 st Law of TD What if we added heat and pushed the piston in at the same time? F

1 st Law of TD Work is done on the gas, heat is added to the gas and the internal energy of the gas increases! Q = W +  U F

1 st Law of TD Some conventions: For the gases perspective: heat added is positive, heat removed is negative. Work done on the gas is positive, work done by the gas is negative. Temperature increase means internal energy change is positive.

1 st Law of TD Example: 25 L of gas is enclosed in a cylinder/piston apparatus at 2 atm of pressure and 300 K. If 100 kg of mass is placed on the piston causing the gas to compress to 20 L at constant pressure. This is done by allowing heat to flow out of the gas. What is the work done on the gas? What is the change in internal energy of the gas? How much heat flowed out of the gas?

P o = 202,600 Pa, V o = m 3, T o = 300 K, P f = 202,600 Pa, V f =0.020 m 3, T f = n = PV/RT. W = -P  V  U = 3/2 nR  T Q = W +  U W =-P  V = -202,600 Pa (0.020 – 0.025)m 3 =1013 J energy added to the gas.  U =3/2 nR  T=1.5(2.03)(8.31)(-60)=-1518 J Q = W +  U = 1013 – 1518 = -505 J heat out