 law of thermodynamics. It states that if to a thermodynamic system if I supply delta q amount of heat, some amount of heat will be utilized to increase the internal energy and part of it will be used to do some work. okay all right so this is the first law of thermodynamics we have learned about the sign convention so I'll write it again if heat is supplied if heat is supplied delta q is positive heat supplied to the system okay otherwise it is negative and then similarly if work done by the system then w is positive otherwise w is negative fine and internal energy is nothing but sum of kinetic energy and potential g so if kinetic energy and potential g increases if kinetic energy plus potential g of molecules increase then delta u is positive otherwise it is negative okay can you tell me a way to increase the kinetic energy of the system by heating it by raising the temperature because kinetic energy of the molecule is nothing but in a way temperature only okay it represents temperature all right and is there any way to change the potential g of the molecules increase the volume not really did we talk about latent heat when the state change happens then kinetic energy doesn't change okay for the molecules kinetic energy remains constant that is why the whenever phase change happened temperature remains constant so what happens when solid converts to liquid potential g of the molecule increases okay similarly when liquid converts to gas or vapor then also potential g increases so in order to increase a potential g increase in temperature is not required all the time okay so energy can be increased by bus just changing the state itself but when we talk about the ideal gas okay this is true for all the objects okay doesn't matter whether it is a gas or solid or liquid anything now if we talk about gas right down if the system if the system is ideal gas okay then potential g is 0 potential g is 0 because potential g is because of the interaction between the molecules okay so only kinetic energy is present that is point number one so potential g is a function of temperature only for the ideal gas okay point number two for the gas we know the work done is force into displacement f into displacement along the direction of force okay and we also have learned that when we talk about fluid we we are more comfortable with dealing with the pressure right so instead of force we are dealing with pressure and for the pressure we have different formulas for example for the ideal gas we have ideal gas equation for the pressure right pv is equal to nrt and for the gas pressure can change with the depth also you can use Pascal's law alright so it is better to write work than in in terms of pressure when we are dealing with gases okay so that is why the force can be written as pressure into area that into displacement okay and I know that the displacement will be always in the directs along the direction of pressure see the pressure is equal in all direction whenever we talk about fluid or gas it is same in all the direction that is what the Pascal's law is okay so suppose there is an expansion going on let's say there is a piston okay there is this piston and gas is inside and when this piston moves up okay the the force because of the pressure will be along the direction of displacement only okay perpendicular to the surface and the displacement direction and the force direction are same that is why area into displacement becomes the change in volume this is suppose the movement of the piston this is let's say s so I can say that the work done is p into whatever extra change in the volume so work done in the case of the ideal gas it is useful to use this law does not matter what is the process and what is happening irrespective of everything you can use this formula directly okay so work done for the ideal gas is given as integral of PDV but that is not true for solids or any other objects so don't confuse PDV integral with any other work done only in case of gas this is valid is it clear to all of you yes sir okay fine let me find out you know some numerical based on this will solve it please do this latent heat of water lyrical vaporization is measured as please write down 2256 joule per gram okay and it is known that one gram of water has one centimeter cube of volume okay now same one gram of vapor at one atmosphere has a volume of 1671 meter not meter cube centimeter cube okay this is what is seen you need to find the increase in internal energy of one gram of water when it converts to one gram of vapor okay this is a direct application of first law of thermodynamics please solve this no one should I wait so do we have to convert the units of work done the 586 no 586 is not correct what do you mean by convert the units of work done everything in joules heat work everything to use joules okay I'll solve this now 2086 is the answer okay 2086 I know that delta q is equal to delta u plus w okay now heat supplied can be written as mass into latent heat of vaporization this is equal to delta u plus work done now work done is what work is done by the system on the atmosphere because the system is expanding from 1 centimeter cube to 1671 centimeter cube the system has expanded so work done is pressure into change in volume of the system delta u will be equal to mass into latent heat of vaporization minus pressure into delta v now when you substitute the values please take care of the SI system now here it is given joule per gram so I can multiply mass in gram so m into Lv will be 2256 only minus pressure that most of the pressure is roughly around 10 to power 5 into delta v delta v is change in volume this minus 1 now it is centimeter cube to convert in meter cube you need to multiply with 10 to minus 6 okay so this will roughly come out to be equal to 2086 joule is it clear any doubts please ask if there is any doubt don't just leave it for the later on okay assuming no doubts I'll move on to the next numerical so here it is that there's an electrical heater that supplies heat at the rate of heat supplied is at the rate of 100 watt and if system does work 75 joules per second please find out at what rate internal energy is increasing 25 yes sir correct so delta q is equal to delta u plus w if I differentiate it with time I'll get dq by dt is du by dt plus dw by dt so dq by dt is rate of heat supplied which is 100 watt is equal to du by dt plus rate at which work is done which is 75 so du by dt is equal to 25 joules per second this is the rate at which internal energy is increasing okay fine so let's proceed forward now that you have put enough hold of what these terms mean