A lab report at uni

Laboratory report: Processes of an ideal gas

Experiment 1
1. Plot a graph showing you experimental results from the instance where you pressurised the vessel to 30kPa. The graph should show:
Time on the x-axis. Pressure (P1) and temperature (T1) on the y-axis
Label all axes and provide a legend to identify each of the data series
Provide the figure with a title in the lower box.

Figure 1: The variation of pressure, P and Temperature, T with time, t.

2. The specific heat ratio of an ideal gas can be determined using the following formula:

Where Ps is the starting pressure (before pressurising the vessel), Pi is the intermediate pressure (stabilized once air pump is switched off) and Pf is the final pressure (stabilized after valve opening). You will need to specify the absolute values at each pressure. Note that the equipment will measure the gauge pressure relative to atmospheric pressure. Therefore you will need to add the atmospheric pressure (take this value to be 100 kPa). Ensure you use a steady stable value for each pressure value. Plotting a graph of your data or analysing the data table should help you identify these values. You might consider taking an average value over a short period of time.
Based on this information complete table 1.
Note: The software does calculate and summarise these pressures in the results spreadsheet but they might not be correct so check!

Run
Ps (kPa)
Pi (kPa)
Pf (kPa)
Specific heat capacity ratio CP/CV
1
100.86
129.24
121.6
1.3258
2
100.72
123.70
118.24
1.2815
3
100.44
119.06
113.42
1.3993
4
100.52
114.46
110.83
1.3301

3. The ideal specific heat ratio of air in the temperature range of your experiment is 1.400. In the box below discuss how this relates to your results and explain why there might be differences between this value and your measured results.

Starting pressure, Ps – The starting pressure should have rather been 0, that is, when the pressure inside the jar is the same as the atmospheric pressure outside it. Then result would have been more accurate.
Intermediate pressure, Pi – it is an approximate value after the air pump has been switched off and therefore it may not be the ideal value to be used. An average value of the intermediate pressure over a certain time could have been used to get a better result.
Final pressure, Pf – this pressure value, again, could have been improved if we extend the lapse of time we used for the experiment as the pressure in the jar would stabilise and almost equally distributed and therefore a more precise result.
Other than the pressure measured, the equipment we used in the experiment could be ‘inaccurate’. This could be due to a loose valve (e.g V1) and/or the jar itself is not well sealed, that is, there may be air leakage at the bottom of the jar. This would lead to pressure loss during experiment

4. In the box below describe the type of process that most closely resembled that which occurred when you rapidly expelled some of the air from the large vessel. In your answer name the process you think occurred, sketch it (diagrammatically) on a pressure-volume diagram showing the direction of the process path. Explain is there was any work done and if so was work done on or by the gas?

When air is expelled through the valve V1, the temperature is observed to decrease. The other variable in this experiment is eventually the pressure inside the jar and also the volume. It should be noted when the pressure is constant, temperature also stays constant. From the graph, a decrease in the pressure of the gas results in a decrease in temperature and an increase in volume. This process is called an adiabatic process as there is no heat supplied to the system although there is change in temperature.

The graph shows when the pressure is decreased by a certain amount ∆P, the volume of the gas increases. Moreover, the temperature of the gas at 1 and 2 are not the same. As a result of which we can conclude it is and adiabatic change process

Relating the first law of thermodynamics to the system:
∆U = Q - W , where ∆U is the change in internal energy, Q is the heat supplied to the system and W is the work done by the system.
Note that we have assumed it is a closed system.
Since there is no heat supplied to the system, Q=0
Therefore ∆U= -W
We can now conclude from the equation above, that work is done by the system.
Apart from the calculations we made above, we could simply think of an expansion (volume increase). For this expansion to occur, work has to be done against atmospheric pressure to increase the volume and also the internal energy of the system decreases.
W = p∆V
∆V = V1 – V2 which is a positive value and therefore showing that work done is also positive and therefore an expansion has occurred. As a result, work has to be done by the gas to produce this expansion.
Hence work is done by system.

Experiment 2
5. The following equation can be used to estimate the ratio of the tank volumes which are 23 litres (tank 1) and 9 litres (tank 2).

Where:

P2absi is the initial absolute pressure of tank 2 in N/m2 ( = Patm –Vi)
P1absi is the initial absolute pressure of tank 1 in N/m2 ( = Patm +Pi)
Pabsf is the final absolute pressure in N/m2

This equation can be derived from the ideal gas equation of state assuming that air behaves as an ideal gas.

Calculate the ratio of volumes from your results. Comment on your answer compared to the actual ratio (23/9) with regard to the magnitude of your result and what might have caused any discrepancy between the measured and actual result.

The value calculated above using experimental values shows that the ratio is much than the expected value 23/9.
This could be explained due to the fact that,
1) A longer time interval would produce a more precise result of the final pressure, as the molecules will be more evenly distributed across the two volumes and will take up all the ‘space’ in the system
2) The final pressure taken is only at one particular time. It is good to note that molecules keeps moving around and at one particular time, there may be more molecules in one jar than the other and therefore the values of pressure keeps on changing. As a result I would suggest that an average of the pressure is done over a certain lapse of time.
3) The absolute pressure of the initial pressure P2 should be 100kPa. However in our measurement, the pressure was higher than the atmospheric pressure. One possible reason to this is, the inner content is being affected by the temperature outside the jar(higher temperature outside will cause a higher pressure inside jar)

6. In order to ascertain the ratio of tank volumes we have used a certain type of process. What was it and why was it used to ascertain the volume ratio? Hint; think about the start and end states of the 2 tanks and the ideal gas equation.
We have assumed that the number of molecules during the process stays the same and the temperature is kept constant through out the experiment. By this way, we can relate the two equations as in the question above using ideal gas equation.
PV=nRT, where R is the universal gas constant.
Since the number of moles, n and the temperature, T are kept constant, the only variables in the experiment are pressure and volume. Therefore equation is
∆P1*V1=∆P2*V2
Rearranging the equation gives us By keeping the temperature constant, the process should be isothermal as in isothermal process, and the variables are pressure and volume, which is the case here. In this experiment, when the volume has been increased, the pressure in the left jar has decreased and that on the right has increased at a constant temperature.

7. What other type of process could be used to ascertain the ratio of tank volumes? Explain your answer. Do the processes you have given as answers to questions 6 and 7 have anything in common?
This time, we could make temperature and volume the variables and keep the pressure constant throughout the experiment. Using the ideal gas equation again, we get,
PV=nRT , where p=constant ,
Note that we should also keep the number of molecules in the jar constant.
This type of process is called an isobaric process.
A possible PV graph would be as shown in the graph here.
In the graph, moving in the direction from 1 to 2, there is an expansion at constant value pressure.
The number of molecules are the same in both answers to questions 6 and 7. In addition to that the volume V1 and V2 are the variables in both answers as this is what we are supposed to find in the experiment and should therefore be the variables and appear in the equation as above.