 A frictionless piston-cylinder device contains 10 pound mass of steam at 60 psi and 320 degrees Fahrenheit. It is now transferred to the steam until the temperature reaches 400 degrees Fahrenheit. If the mass is constant, determine the work done by the steam during the process. So I will draw this piston-cylinder arrangement and recognize that if heat is transferred to the steam, that's going to be a Q in, a heat transfer in, then heat is transferred to the steam, and in order for the steam to remain at a constant pressure while receiving heat and increasing in temperature, the piston is going to have to move up. We are going to have to expand in this process. Think about it. As you heat up the steam and the molecules are moving faster, they're bumping against the walls a lot more, there is going to be a pressure increase associated with that increase in temperature. So in order for us to maintain a constant pressure, we are going to have to have the piston move up. Now again, we find ourselves in a situation where we deduce that via context clues. Now I knew that this was an isobaric process because of the words frictionless piston-cylinder device and this diagram that indicates the piston as floating on top of the substance. If the piston is floating on top of our substance and we have no friction, then the piston will provide the same weight across the same area the entire time. It can move up and down, but its weight is the same and its area is the same, therefore its pressure is the same. Those are the context clues that allow us to deduce that this is an isobaric process. So we have expansion from one to two. That's going to be a positive change in volume and I'm going to have a boundary work associated with it. Since boundary work is defined as the integral of pressure with respect to volume and my pressure is constant, this is going to become pressure times the integral of the volume from v1 to v2, which is just going to be pressure times v2 minus v1. Now I know that the steam is initially at 60 psi and since it's an isobaric process that means the pressure at the end of the process is also 60 psi, therefore the pressure that I'm going to multiply by in this equation is 60 psi. Now do I know the volume? I don't, but I do know two independent intensive properties and remember that it takes two independent intensive properties to fully define a state point. At state 1 I know the pressure and the temperature. Those two independent intensive properties allow me to come up with any other intensive property that I need, including but not limited to specific volume because remember total volume is not intensive and at state 2 I know p2 is equal to p1 which I know and I know t2 therefore I could come up with whatever I want including but not limited to v2. That specific volume can be determined knowing pressure and temperature. Now right now in chapter 2 the only skill set that we have for determining a specific volume from a temperature and pressure would be to use the ideal gas law but eventually we will learn how to use the property lookup tables and we will be able to look this up for steam given a temperature and pressure. But since we're still in chapter 2 I have done that for you. In the bottom quarter of this example problem we have the specific volume of steam at 60 psi and 320 degrees Fahrenheit and we have the specific volume of steam at 60 psi and 400 degrees Fahrenheit. So we know v1 and v2 and again that's just a stand-in for your eventual ability to look this up in the property tables. Whatever the case we know specific volume 1 and we know specific volume 2. Therefore I need to describe the total volume in terms of specific volume. Well remember that specific volume is total volume divided by mass therefore total volume could be represented as mass times specific volume. Therefore I could write this as pressure times mass 2 times specific volume 2 minus mass 1 times specific volume 1 and I can reasonably assume that the same 10 pound mass of steam is present through the entire process. It's reasonable to assume that no steam leaks around our piston so I'll write that as an assumption mass is constant i.e. closed system in which case the mass can be factored out and I have mass times pressure times little v2 minus little v1. I know both specific volumes because we quote looked them up unquote I know the mass and I know the pressure so I can compute an answer. So the boundary work is going to be 10 pound mass multiplied by 60 psi by the way I was just told 60 psi. Should we treat this as a gauge pressure or an absolute pressure? Well the general rule of thumb is if you have no indication of it being a gauge pressure then you should assume it's an absolute pressure. An example of the type of indication that you would see would be pressure is right off of a gauge on cylinder says 60 in that case we would call it gauge pressure or if we had the weight of the piston and we had taken divided by the area that would be a gauge pressure because in that analysis we weren't including atmosphere pressure. Whatever the case though let's assume it's an absolute pressure. You will commonly see this as 60 psi a to indicate absolute pressure and psi g to indicate gauge pressure. Anyway 60 psi and then I'm multiplying by the difference between 0.5216 0.5216 and 0.4 something that was wrong page 0.4674 4674 4674 and that quantity was in cubic feet per no it's in cubic meters per kilogram. That's interesting I mean presumably we would have looked them up on the imperial tables instead of the metric tables but you know what this is an excellent opportunity for some practice at unit conversions in our calculations and what unit did we want presumably BTUs because everything else is no kilojoules that's fun so we have a mixture of imperial and metric units and we are going to calculate a work in kilojoules so I will start as is good practice with my destination and work backwards furthermore I'm probably going to need more space so I will spread out a little bit here. So a kilojoule is a thousand joules and a joule is defined as a Newton times a meter and a Newton is a kilogram meter per second squared and we don't necessarily have to go all the way back because we have a pound of force and a pound of mass here so we could presumably leave this in Newtons and at that point convert from pounds of force to Newtons or you could break the pound of force apart into its pound mass components it's entirely up to you since I have pounds of force already I will leave it in Newtons so that those force dimensions will cancel again you're starting on a secondary dimension you're working backwards and still until stuff cancels you don't necessarily have to go back all the way every time so joule cancels joules my mass dimension will eventually cancel mass dimension I will have two more meters that will appear once I break out the psi into pounds of force per square inch psi is a pound of force per square inch so length dimension in the numerator to the third power and then we have meters and square inches in the denominator which is still length times length times length those will cancel and then again the pound of force and the Newton can be converted and the kilogram and the pound of mass can be converted so for that we will go to our conversion factors sheet and it says that a kilogram is 2.2046 pounds of mass that kilogram was in the denominator so I want it in the numerator and that was 2.2046 pound mass by the way how did I know that it was a pound of mass because it's appearing under the mass and density also your textbook uses just lb for pound mass and lbf for pound of force so if it isn't labeled it's a mass then we need inches to meters inverted so I will say that a meter is 3.2 808 feet so one meter is 3.2 808 feet and then apparently I will change pages again and then a feet is 12 inches and I want my inches squared to be in the numerator so I will actually come at that the opposite direction one meter in the denominator and then 3.2 808 feet and then one feet is 12 inches and then I will square everything because I want to get rid of inches squared so squared squared squared squared squared squared so something went wrong with my screen capture there I'm not exactly sure how much of that you missed so I'm just going to jump all the way back to this step and we'll walk through that one more time just to make sure that we're all on the same page so there were our goal is to get inches squared in the numerator which means feet has to be in the denominator and then feet has to be in the numerator therefore meters has to be in the denominator there are one meter in 3.2 808 feet and there is going to be 12 inches in each of those feet and I square everything because I want to get rid of inches squared one square is boring one square is boring meters squared is going to cancel nothing yet oh yeah cubic meters cubic meters is going to cancel square meters and meters and then a square feet cancel square feet inches squared cancels inches squared so I got rid of those length dimensions and then kilogram cancels kilograms pound of mass cancels pound of mass so all I need is the conversion from pounds of force to newtons for which I go back to my conversion sheet I scroll on down to force and under force I see that one newton is 0.2 248 pounds of force but I like to write the number that's bigger than one so I don't confuse my decimal places so I will say one pound of force is 4.4482 newtons one pound of force is 4.4482 newtons one pound of force 4.4482 newtons and then the pound of force in psi cancels the newtons and the newtons cancel the newtons that leaves us with kilojoules so calculator if you would join us we can compute a result we have 10 actually I'll put that in parentheses 10 times 60 times the quantity 0.5216 minus 0.4674 times 3.2808 squared times 12 squared times 4.4482 and then I'm dividing that entire quantity by 1000 times 2.2046 and then one squared one squared and one okay so we get 101.7 kilojoules so our boundary work is a positive quantity which implies expansion because remember that when we built our equation for boundary work our dv term represents a positive change in volume therefore a positive boundary work implies expansion and that is going to be the work done by the system so the system is applying the work if it had been a compression process that would be worked on on the steam not worked on by the steam that distinction is important especially when we start bringing different terms together into the energy balance for good measure I like to call them inputs and outputs work done by the steam would be an output work done on the steam as an input therefore map this somewhere in your brain a positive boundary work is a work out a negative boundary work is a work in got it so this is a work out of 101.7 kilojoules which sounds like a radio station