 Let us now come to the question number 2, wherein again we have been given two properties, but importantly now we have got mass of 2 kg here that has to be not forgotten because in actual problems many times the masses could be different. We have got P is equal to 15 MPa here and the property tells us that it is a saturated vapor. That means I have to refer to Vg, Ug, Hg and Sg that kind of a thing I should go to 15 MPa straight away. Important to realize that the moment I say saturated vapor or saturated liquid I will have to go to either Vf or Vg respectively. So, let us go to now table number 2 where the 15 MPa we can trace and get the values of Vg, Ug, Hg and Sg and that is the answer to this question, but those values will be per kg what you have to do because we want to have at 2 kg we will have to multiply those values by 2 kg and that will be the answer. So, let us go to the table number 2 and find out the values at 15 MPa. Here we are so this is 15 MPa point here and correspond to that what you have are these different values. Vf, Vg, Uf, Ug and what we are going to take basically because we know that this saturated vapor we are going to take all the values related to Vg, Ug, Hg and Sg. So, I will come back and write down those values. So, question 2 M is equal to 2 kg P is equal to 15 MPa and condition is saturated vapor. We just found those values therefore V is equal to Vg which is 0.01033 meter cube per kg from the table and in actual case now we got 2 kg and therefore in our case Vg will be 2 times this 0.020676 meter cube. Similarly now Ug is equal to 2455.8 kilo joule per kg in our case Ug will be equal to 4911.6 kilo joule alright not per kg now. Similarly Hg is from the table is 2610.67 kilo joule per kg in actual case our case Hg will be equal to 521.4 kilo joule and Sg is equal to 5.31106 kilo joule per kg which in our case for 2 kg it will be S is equal to 10.6212 kilo joule per Kelvin. So, this was kilo joule per kg Kelvin this will be kilo joule per Kelvin now and this is the value of Hg. So, these are our values for 2 kgs alright. Let us put those values in the answer sheet. So, second problem we get all those values here 0.02 meter cube 4911 521 10.6212 that is what the values would be in this case. Now, let us come to the last problem. This problem is little tricky what you have here is R2 conditions but the conditions are T is equal to 170 degree centigrade that means temperature is given and the second property that has been given is enthalpy 2816.2 alright. So, these 2 values are given and now based on this I would like to find out all other properties that means V, U and S. One can always find out pressure also from this alright. So, let us find out these properties for which we will have to go basically from let us go to table 1 and find out in which zone this lies first alright because from T is equal to 170 degree centigrade from table 1 we can find out what is Hf and what is Hg and then we will compare the given H value and then find out in which zone this system lies. So, what we can do let us go to table 1 locate T is equal to 170 degree centigrade and find out what are the values of Hf and Hg are alright. T is equal to 170 degree centigrade we can go to the enthalpy part and we can see that Hf and Hg are these values alright. Just note down these 2 values of Hf and Hg. So, let us come back to this problem solving now. So, having seen these values of enthalpy at T is equal to 170 degree centigrade now let us compare these values with the given H value in the problem number 3 in this case. So, what is our question 3? T is equal to 170 degree centigrade and H is equal to 2816.2 and we have just found from table 1 at T is equal to 170 degree centigrade Hf is equal to 719.08 kilojoule per kg and Hg is equal to 2767.9 kilojoule per kg. It means that H given is more than Hg because H given is 2816.2 while Hg is 2767 it means that we are in superheated zone. So, having established that we are in superheated zone we should now go to table number 3. The problem now here is we just know that it is in superheated zone but we just know temperature we do not know the pressure. When in table number 3 we have got different tables at different different pressures. So, what I am going to do now is by trial and error I will go through every pressure and see at T is equal to 170 degree centigrade what is the value of H and then I will get a feel at what pressure it should lie. So, it is a little bit reverse exercise that we are doing that we know the value of H we do not know the pressure we just know the temperature. So, I will go to different pressures now and get the corresponding enthalpy at T is equal to 170 degree centigrade. So, let us go to 0.01 MPa first in table 3 and see what is the value of H at T is equal to 170 degree centigrade. So, here we are at table number 3 P is equal to 0.01 MPa and let us locate T is equal to 170 degree centigrade here on this. So, this is 170 degree centigrade and corresponding value of enthalpy is 2821. So, just note down the value at first at 0.01 MPa then we can look at the values for the next table. Let us see at 0.02 MPa and let us see at T is equal to 170 degree centigrade. So, here we see that the enthalpy is 28209 that means the enthalpy is reducing the H value that we have is 2816.2. That means as we went from 0.01 to 0.02 the enthalpy had reduced down. So, we have to go further down and we can see for 0.05 MPa for example and let us go down and write down those values. Look at this table now P is equal to 0.05 MPa and T is equal to 170 degree centigrade. So, I can see 2819.2. So, I should go further down and at P is equal to 0.1 MPa and T is equal to 170 degree centigrade I see 2816.2 that means we were at pressure P is equal to 0.1 MPa. Please note all these values of V, U, H and S accordingly and that is our answer for the table that we are going to write. So, I will now go to go back to our table. This was the condition this is 1 kg T is equal to 170 degree centigrade what was given to us was enthalpy H 2816.2 what we have found out from here are the other values. We found that 0.1 MPa and this we did by trial and error we have got the values of V, U and S at this T value and correspondingly what we had was enthalpy 2816.2 which was a region in the superheated zone. So, thank you very much.