For question 3C, why did we convert n in n deltaHvap to m(1mol/18.02g)?
I'm confused as to why they decided to change it.
Winter 2014 Midterm Q3C
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AnkitaNair1E
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Re: Winter 2014 Midterm Q3C
Hi Amy,
To answer your question, I think they decided to substitute the delta n value with (77.0408g/18.02g) because that is the change in moles of steam. Essentially what is happening is that as the water gets heated and converts to steam, the steam does work on its surroundings. We know that the work equation is -P(delta v) because the problem doesn't tell us that the system is reversible or isothermal. Also since we don't know the pressure or the change in volume but we know that pv=nrt we can substitute to get work=-(delta n) RT. We know what R and T are but we need to find delta n. One way to do this is to understand that originally problem starts off with 77.0408 grams of hot water and 0 grams of steam. By the end of the problem, all of the water is converted to steam so the change in the amount of steam is 77.0408 g (because mass final -m initial= (change in mass)). Now that we know the change in mass, we can divide it by the molar mass to get the change in moles (i.e. 77.048/18.02). by doing so we get (delta)n and can then solve the equation to get the amount of work done by the steam.
As you may have noticed, the course reader says that they would have accepted w=0. This is because the problem is somewhat poorly worded and forgets to tell us whether or not the volume is changing. If we assume the volume changes, we do the steps above to find work. If we assume that the volume is constant, then work would be zero because -p* (delta)V would =0.
Normally when you get a problem like this, they should tell you if the process is reversible isothermal (so you know what work equation to use) and if the volume or pressure stays constant. If they don't give you specific information its safe to assume you can use p*(delta)V. This lack of information just made the question a bit more tricky and required you to make assumptions that could have led you to get different answers.
Hope this helps!
To answer your question, I think they decided to substitute the delta n value with (77.0408g/18.02g) because that is the change in moles of steam. Essentially what is happening is that as the water gets heated and converts to steam, the steam does work on its surroundings. We know that the work equation is -P(delta v) because the problem doesn't tell us that the system is reversible or isothermal. Also since we don't know the pressure or the change in volume but we know that pv=nrt we can substitute to get work=-(delta n) RT. We know what R and T are but we need to find delta n. One way to do this is to understand that originally problem starts off with 77.0408 grams of hot water and 0 grams of steam. By the end of the problem, all of the water is converted to steam so the change in the amount of steam is 77.0408 g (because mass final -m initial= (change in mass)). Now that we know the change in mass, we can divide it by the molar mass to get the change in moles (i.e. 77.048/18.02). by doing so we get (delta)n and can then solve the equation to get the amount of work done by the steam.
As you may have noticed, the course reader says that they would have accepted w=0. This is because the problem is somewhat poorly worded and forgets to tell us whether or not the volume is changing. If we assume the volume changes, we do the steps above to find work. If we assume that the volume is constant, then work would be zero because -p* (delta)V would =0.
Normally when you get a problem like this, they should tell you if the process is reversible isothermal (so you know what work equation to use) and if the volume or pressure stays constant. If they don't give you specific information its safe to assume you can use p*(delta)V. This lack of information just made the question a bit more tricky and required you to make assumptions that could have led you to get different answers.
Hope this helps!
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