Work done by a heat engine - quick issue with units used in the solution

In summary, the lecturer made a schoolboy mistake by using litres in his calculations and this caused the total work done by his calculations to be much higher than if he'd used decimetres.
  • #1
Faloren
12
0

Homework Statement



"A gas storage system contains oxygen at a pressure of 5 atmospheres at 300 K.
(Here, you may assume that oxygen behaves as an ideal gas.) The volume of the
cylinder is initially 150 L. (State 1.) In a fire the gas is initially heated by 200 K,
with a safety system keeping the pressure constant (state 2). A sprinkler system
then reduces the temperature to 280 K but the volume does not change because the
safety mechanism has jammed (state 3). The gas is then slowly compressed back to
150 L as the safety mechanism unsticks with the temperature staying constant (state
4). Finally, the sprinklers stop and the system warms up at constant volume to
300 K so that it is back in state 1."

(i) Sketch these processes on a pV diagram showing the numerical values of
p, V and T in each of the 4 states. {5}
(ii) What is the net work done by the oxygen? (You may assume that Cp and Cv
are independent of temperature.) {5}

Homework Equations



pV=nRT

equations for work with isothermal processes (don't bother with the constant pressure, it's just the area of the rectangle)

i.e W = nRT ln (V2/V1)

The Attempt at a Solution



Now, I'm happy about the first part (i plotted the diagram correctly). I did forget that i could use the relationships that P1/T1 = P2/T2 etc, but i used the ideal gas equation to work out all the bits individually and they were mostly the same as the answer scheme (only one rounding error - 4.71 calculated against 4.67 exact).

However, the lecturer does some really funny things on his mark scheme. First and foremost, he uses litres in all his volume calculations.

I know all the data from state 1 to work out the number of moles of oxygen in the system, so i said that:

V = 150L = 0.15m^3, P = 5atm = 5*1.01*10^5 Pa, T = 300K given

Plugging this into the ideal gas equation gives just over 30 moles by my calculation. However, when i went back to mark the exam paper with the supplied solutions, i noticed that the lecturer had used litres in almost all his workings - giving far more moles of oxygen than i got (factor of 1000 roughly). This also meant that the total work done by his calculations was something like 15GJ, which seems ludicrously high. It just seems odd that'd he'd make such a "schoolboy" error, so i wondered whether i was at fault. In theory the results should be the same, regardless of what volume measurement you use (and as far as i understand the volume of a litre of substance doesn't depend on the substance!)?

Please note that I'm not concerned with the actual answer, I'm confident that my working is sound even if the numbers are off.

Question: Am i right to say that you must convert to SI units before plugging into the ideal gas equation (i.e. metres rather than litres)?
 
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  • #2
Faloren said:
Question: Am i right to say that you must convert to SI units before plugging into the ideal gas equation (i.e. metres rather than litres)?

Yes. Everything must be consistent. If everything else is in SI units, your volume should be in SI units. A litre is a cubic decimetre. So you could use litres IF you converted all of your other units to a system that uses a decimetre as the unit of length. Otherwise, no.
 
  • #3
Thanks :)

Hah i really hope he didn't mark everyone down last year because of that!
 

Related to Work done by a heat engine - quick issue with units used in the solution

1. What is the definition of "work done by a heat engine"?

The work done by a heat engine refers to the amount of energy that is converted into mechanical work by the engine. It is typically measured in units of joules (J) or foot-pounds (ft-lbf).

2. How is the work done by a heat engine calculated?

The work done by a heat engine is calculated using the formula W = Qh - Qc, where W is the work done, Qh is the heat supplied to the engine, and Qc is the heat rejected by the engine.

3. Why is it important to consider the units used in the solution for work done by a heat engine?

It is important to consider the units used in the solution for work done by a heat engine because they can affect the accuracy and consistency of the calculation. Using the wrong units can lead to incorrect results.

4. What are the commonly used units for work done by a heat engine?

The most commonly used units for work done by a heat engine are joules (J) and foot-pounds (ft-lbf). However, other units such as kilojoules (kJ) and British thermal units (BTU) may also be used.

5. How can I convert between different units for work done by a heat engine?

To convert between different units for work done by a heat engine, you can use conversion factors or conversion calculators. It is important to ensure that the units are converted accurately to avoid any errors in calculations.

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