 So, let us now come to the question B of this tutorial 4 which is now relating to the entire water phase diagram to the first law of thermodynamics which you have already learnt and let us try to apply this knowledge what we have learnt from the phase diagram to the first law and where we do the energy and work balance and try to find out the phases at various locations in the phase changes or the property changes. So, let us come to the problem and question B says now compute the changes in the thermal energy and enthalpy and expansion work done when 1 kg of saturated liquid water at T is equal to 100 degree centigrade evaporates isothermally to dry saturated steam. So, one can understand from here that initially the it was a saturated liquid and the final state it saturated steam or dry saturated steam that means the entire process happens at the same pressure and the same temperature when 100 percent of liquid gets converted to 100 percent of dry saturated steam. So, can you guess from here where does the state 1 and where does the state 2 lie all right let us now come to the solution of this problem. So, for understanding we can take a cylinder and piston arrangement where in we can have saturated liquid in the volume and once the liquid gets evaporate to steam we can get work done from this system. So, we can take a cylinder piston arrangement to understand this problem. So, our system could be a cylinder piston arrangement consist of cylinder piston arrangement and this is this dotted line shows my system all right. Also we can draw the diagram on a PV process diagram on a PV plane. So, this is P and V and we can understand from the question that my initial state it as saturated liquid region the final state is on saturated vapor region these are 0.1 and 0.2 and this process happens at constant pressure and constant temperature and this P happens to be we have been given temperature which is T is equal to 100 degree centigrade all right. That means we should refer to basically table 1. So, what do we have to refer we go to table 1 look at 100 degree centigrade and get the values at the f point and the g point because the initial points lies on a saturated liquid curve and the final point lies on a saturated vapor curve. So, we get the f values and the g values and we can solve the problems. So, if I go to the steam table now and get the values at 0.1 state 1 and state 2 this is state 1 and this is state 2. So, can we get to the steam table? So, this is stable number 1 and 100 degree centigrade. So, we know 100 degree centigrade is the phase change point that means we have to be at 0.1 MPa that is atmospheric pressure and corresponding to this now we can look at get all the properties related to VF, VG, UF, UG, HF, HG and SF, SG. Please note down all these values then we can come back to solution again. So, the properties at state 1 and state 2 can be read from the steam table and I will write down those values. So, VF is equal to 0.00104 meter cube per kg and for state 2 VG is equal to 1.67 meter cube per kg. Then I will get UF is equal to 419.06 kilojoule per kg and I will get UG is equal to 2506 kilojoule per kg. The third property I am concerned about is enthalpy which is HF is equal to 419.17 and HG is equal to 2675.6 kilojoule per kg. So, these are different properties that we read for state 1 and state 2 as shown in the diagram. Now, we understand that this process expansion the process of expansion is a constant pressure process alright. So, expansion work W expansion is equal to because this is a constant pressure process P is equal to V2 minus V1 into M and V2 is nothing but VG, V1 is nothing but VF. If we write those values the pressure is 0.1 MPH which is 0.10142 into 10 to the power 6 per Newton per meter square into V2 minus V1 that is our VG minus VF here 1.6718 minus 0.00104 meter cube per kg into 1 kg. This will give us expansion work which is equal to 169448.13 joules which is equal to 169.44 kilojoules. So, this is my expansion work we have got different properties this was one of the questions that was asked the other question is what are delta U and what are delta what is delta H. So, delta U is nothing but change in thermal energy which we have just calculated we have seen that M into UG minus UF which is equal to if I put these values of UG and UF here I will get around 2086.94 kilojoule which is nothing but actually UFG because this is per kg. Similarly, delta H is nothing but M into HG minus HF which is nothing but M into HFG is not it HFG which is equal to 2256.4. So, my answers therefore are the expansion work is equal to 169.44 kilojoule and the change in thermal energy and the change in enthalpy is given by this formulation alright. So, this is the answer for this question let us come and write those answers alright. So, change in enthalpy is 2256.4 kilojoule change in thermal energy delta U which is 2086.94 kilojoule and work of expansion is 169.45 kilojoule which is a positive work done by the system alright. So, this is the first problem that we have seen where we apply the water basically water substance use the steam table and apply to the first law of thermodynamics. Thank you.