**Outcomes** - First Law of Thermodynamics - Able to relate work done with pressure - Understand what happens in systems when work done is negative or positive - Calculate the change of heat, change of internal energy, and the work done using the energy balance equation **Thermodynamics** - Study of conversion between heat and mechanical work - Study of pressure, temperature and volume of matter - Conservation of energy principle **Laws** - [[Engineering Principles/Heat#Zeroth Law of Thermodynamics|Zeroth Law]] - First Law deals with conservation of energy - Second law is the law of 'disorder' and deals with the efficiency of energy conversions - The Third Law concerns matters related to the absolute zero of temperature - Impossible to reach the temperature of absolute zero - There is no fundamental way of deriving them from other theories ## First Law of Thermodynamics - First law is a statement of the law of conservation of energy in a system which converts energy between heat and work - The total energy of a system is equal to the difference between the total heat added to the system and the total work done by the system $ \Delta Q=\Delta U+\Delta W $ Where: * $Q$ is the heat energy * $W$ is an amount of mechanical work * $U$ is the internal energy stored This tells us that if we add a small amount of heat energy to a system, we will either: * Increase its internal energy, OR * some work will be done. OR * a combination of the two The sum of the increase in internal energy and the work done will always be exactly equal to the amount of heat supplied. Either of these may be negative in a particular case. Consider a tube of gas at pressure p closed by a piston of area A. * Suppose the gas is allowed to expand a little, without a significant change in pressure. * The piston moves out by a small distance $\Delta x$. To hold the piston in place, an equal and opposite force must be applied externally. * The force exerted by a gas on the piston: F=PA. The work done by the gas against the restraining force F is: $ \Delta W=F\cdot\Delta x = PA\cdot\Delta x $ But $A\cdot\Delta x=\Delta V$, therefore $\Delta W=PA\cdot\Delta x=P\cdot\Delta V$ Example: if the volume increases from 0.1m^3 to 0.102m^3 at the pressure of 10^5 Pa, then the work done by the gas is: $ \Delta W=P\Delta V=10^5\cdot(0.102-0.1)=200J $ The gas does 200J of work, which is energy transferred to something else (e.g the piston). If the volume decreases, we can say that the gas does negative work. If the volume decreases from 0.1 to 0.098, then the work done to compress the gas is: $ \Delta W=P\Delta V=10^5\cdot(0.098-0.1)=-200J $ In general: * Work done by a gas on its surroundings is positive **(expansion)** * Work done to/on a gas by its surroundings is negative (**compression)** ## Internal energy Kinetic theory shows that heat energy stored within a perfect gas is the kinetic energy of the molecules. $ KE=\frac{1}{2}mc^2=\frac{3}{2}nRT $ Where: * $m$ is the mass of one gas molecule * $c$ is molecular speed * $n$ is number of molecules of gas * $R$ is the universal gas constant * $T$ is the absolute temperature This energy is known as internal energy, symbol $U$. We do not usually need to know the exact value of U, generally we are only interested in how it changes. A change in internal energy always implies a change in temperature in the same direction (in the absence of a change of phase). ## Energy conversion $ \Delta Q=\Delta U+\Delta W $ For gas under constant pressure, $\Delta W=p\Delta V$, so: $ \Delta Q=\Delta U+p\Delta V $ This, if an amount of heat $\Delta Q$ is supplied to the gas, then it must either: * Be stored as an increase in energy * Make the gas do work (ie expand) * Or a combination of the two Negative values mean an energy flow the other way: heat removed, temperature falls, gas compressed. ## Example 1 In a two-stroke steam engine, steam at 5 bar acts on a piston of area $25cm^2$ and moves 6cm per stroke. Given: * Pressure = 5 bar * Area = $25 cm^2$ * $\Delta x$ = $6cm$ How much work is done in one stroke: $ \Delta W=F\cdot\Delta x=pA\cdot\Delta x $ $ \Delta W=pA\cdot\Delta x=(5\times 10^5)\times(25\times 10^-4)\times(6\times 10^{-2}) $ $ \Delta W=75J $ ## Example 3 If 15J of heat are supplied to a gas, and the gas does 20J of work, how does the internal energy of the gas change? $ \Delta Q=\Delta U+\Delta W $ $ \Delta U=\Delta Q-\Delta W = 15-20=-5J $