MECH 351 Lab 1

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Lab Report on Experiment 1 EXPANSION OF A PERFECT GAS MECH-351 Lab Section WI-X Group B January 23 rd , 2024 Dr. Lyes Kadem & Dr. Charles Kiyanda Concordia University Montreal, QC, Canada

OBJECTIVE This lab contains 3 distinct experiments with different objectives which all use temperature and pressure sensors [1]. Experiment 1 aims to show the relation between temperature and pressure by using a short depressurization period [1]. Experiment 2 aims to use the same short depressurization technique to simulate a reversible, adiabatic expansion, which will then be used to calculate the specific heat capacity ratio of the surrounding air through pressure readings [1]. Experiment 3 aims to use a longer depressurization period to simulate an irreversible, adiabatic expansion which will then be used to calculate the specific heat capacity ratio of air by using the pressure readings [1]. INTRODUCTION The behavior of a gas is difficult to perfectly predict, or calculate under most conditions, as there are numerous parameters to take into account. To avoid measuring and calculating many of these parameters, one can use the ideal gas equation along with the assumption that the gas behaves ideally to calculate approximate values for the gas’ state [1]. A gas that behaves ideally is referred to as a perfect gas and the four following assumptions are made about it: gas particles hold negligible volume, gas particles are equally sized and have no intermolecular forces, gas particles move in accordance to Newton’s Laws of Motion, and gas particle experience no energy loss due to perfect elastic collision [2]. Using these assumptions, one can now relate the pressure, temperature, and volume of the gas through the following equation referred to as the ideal gas equation: PV = nRT where “P” is pressure in kPa, “V” is volume in m 3 , “n” is number of moles, “R” is the gas constant in kJ/(mol*K), and “T” is the temperature in K [1]. A real gas can be described by: PV = znRT where “z” is the compressibility factor which has no units [1]. The compressibility factor remains close to one when pressures are below 300 kPa, and thus a real gas can be said to behave ideally under those conditions [1]. The amount of energy required to increase one unit mass by one degree is referred to as the specific heat capacity [1]. The specific heat capacity of a gas varies based on whether the gas is at constant volume, or constant pressure [1]. These are referred to as “C V ” and “C P ” respectively. By taking the ratio of C P /C V , the specific heat capacity ratio, “ ” 𝘠 can be found [1].

This new variable has no units and allows one to relate pressure, temperature and volume during simple expansion or compression of gasses [3]. More applications of this variable exist, but this experiment focuses on the compression and expansion of air. An isothermal process is one in which the temperature remains constant, while an adiabatic process means that no heat energy is added or removed during the process [1]. An isothermal process is not necessarily adiabatic as a system can remain the same temperature if equivalent energy is lost through work as is added through heat [1]. Reversible and irreversible processes refer to the ability of the system to return to its original state with no change to its surrounding environment through heat or work [1]. Processes tend to be irreversible as they often encompass a change in entropy [1]. A process is only said to be reversible if there is no change in entropy due to the process [1]. The following formula is used to calculate the change in entropy for processes done on ideal gasses: s 2 - s 1 = C V ln(T 2 /T 1 ) + Rln(V 2 /V 1 ) where “s” is the entropy of the gas [1]. To achieve such a reversible process, a technique named “Clément Desormes Method” will be employed [1]. This two step method takes an initial state and puts it through a reversible, adiabatic process to an intermediate state [1]. The intermediate state is then left at constant volume until it has stabilized [1]. This process allows one to calculate the specific heat capacity ratio through the following equation: Figure 1: Equation 1.13 [1] The same method can also be applied to an irreversible, adiabatic process and yields the following formula: Figure 2:Equation 1.17 [1]

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This experiment uses a pressurized vessel with internal pressure and temperature sensors, along with software which allows one to visualize the values of the pressure and/or temperature [1]. The vessel has a constant volume and the change in mass is negligible, and therefore treated as a constant [1]. This means that the pressure and temperature are expected to be directly proportional, according to the ideal gas equation. Using this relationship, a reversible and non- reversible, adiabatic process will be done to explore different aspects of the first and second laws of thermodynamics [1]. PROCEDURE Equipment Information The three experiments for lab 1 use an apparatus called “The Expansion of a Perfect Gas TD1004V” [1]. This machine consists of two pressurized vessels, a pressure and a vacuum vessel, with pressure and temperature sensors, along with a connecting bore pipe fitted with ball valves to either transfer, or empty the gasses in the vessels from the top [1]. Both vessels can have a small tube with a valve connected through the bottom to allow the connection of an air pump to control the pressure within the vessels [1]. Below, Figure 1 will show the full apparatus used for the three experiments: Figure 3: Front View of Expansion of a Perfect Gas Apparatus [1]

The three experiments use the pressure vessel, along with valve 1, in conjunction with a diaphragm pump to increase pressure within the vessel [1]. The pump can be controlled via the apparatus’ control box where the various sensors are also connected to [1]. The control box can be seen below in Figure 3: Figure 4: Expansion of a Perfect Gas Apparatus Control Box [1] Various configurations of how the sensors are plugged will have their own distinct advantages and disadvantages which will be briefly seen in the experiments [1]. The final equipment is a software named Versatile Data Acquisition System (VDAS ®) [1]. The software receives the data from the sensors and plots it so that the user may visualize and use the data [1]. Experiment 1 Run VDAS ® and select “Experiment 1” on the computer [1]. Ensure that the pressure vessel is selected and input the ambient pressure [1]. From Figure 1, insert “P1” and “T1” into Figure 2’s “P” and “T” slots under the “Fast Inputs” label [1]. Connect both “Pressure Ports” from Figures 1 and 2 and start communications on VDAS ® [1]. Turn on the air pump, let the pressure rise to approximately 30kPa, and then turn off the air pump [1]. Wait for the temperature and pressure to stabilize within +/- 0.5 and kPa, and then quickly open and close ℃ Figure 1’s “V1” valve [1]. Wait for the temperature and pressure to stabilize within +/- 0.5 ℃ and kPa, and stop VDAS ® communications [1]. Scroll through the acquired data graph until the offset is found and right click the graph and select “Save analogue traces to File” [1]. Ensure that a copy of this file is acquired for use in the lab report and then open “V1” to completely empty the vessel [1].

Experiment 2 Run VDAS ® and select “Experiment 2” on the computer. Ensure that the pressure vessel is selected and input the ambient pressure [1]. From Figure 1, insert “P1” and “T1” into Figure 2’s “P” and “T1” [1]. Connect both “Pressure Ports” from Figures 1 and 2 and start communications on VDAS ® [1]. Run the air pump until the pressure reaches approximately 30KPa and then stop the air pump [1]. Wait for the pressure and temperature to stabilize within +/- 0.5 and kPa, and record the pressure in the allotted spot on the data sheet [1]. Quickly ℃ open and close valve “V1” and then allow pressure and temperature to stabilize within +/- 0.5 ℃ and kPa, [1]. Stop VDAS ® communications and record the final pressure in the allotted spot on the data sheet [1]. Scroll back on the data graph until the offset is reached and record the lowest pressure into its allotted spot on the data sheet [1]. Right click the graph and select “Save analogue traces to File” [1]. Ensure that a copy of this file is acquired for use in the lab report and then open “V1” to completely empty the vessel [1]. Repeat the entire process, excluding saving and copying the file, two more times, and then another 3 times each for 40kPa, 50kPa, and 60kPa [1]. Experiment 3 Run VDAS ® and select “Experiment 2” on the computer. Ensure that the pressure vessel is selected and input the ambient pressure [1]. From Figure 1, insert “P1” and “T1” into Figure 2’s “P” and “T1” [1]. Connect both “Pressure Ports” from Figures 1 and 2 and start communications on VDAS ® [1]. Run the air pump until the pressure reaches approximately 30KPa and then stop the air pump [1]. Wait for the pressure and temperature to stabilize within +/- 0.5 and kPa, and record the pressure in the allotted spot on the data sheet [1]. Open valve ℃ “V1” and then close after 1 second then allow pressure and temperature to stabilize within +/- 0.5 and kPa, [1]. Stop ℃ VDAS ® communications and record the final pressure in the allotted spot on the data sheet [1]. Scroll back on the data graph until the offset is reached and record the lowest pressure into its allotted spot on the data sheet [1]. Right click the graph and select “Save analogue traces to File” [1]. Ensure that a copy of this file is acquired for use in the lab report and then open “V1” to completely empty the vessel [1]. Repeat the process two more times, excluding the saving of the file [1].

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RESULTS Experiment 1 Figure 5: Temperature (top blue line) and pressure (bottom yellow line) readings from Experiment 1 on “P vs t” and “T vs t” Graph

Experiment 2 Sample Calculations Formulas used: P abs = P atm +P gauge where P atm is 102 kPa and Figure 6: Equation 1.13 [1] P abs = 102 kPa + 31.79 kPa = 133.79 kPa rev= [ln(119.47 kPa) - ln(133.79 kPa)] / [ln(123.3 kPa) - ln(133.79 kPa)] = 1.386 𝘠

0.3 bars 0.4 bars 0.5 bars 0.6 bars Trial 1 Trial 2 Trial 3 Trial 1 Trial 2 Trial 3 Trial 1 Trial 2 Trial 3 Trial 1 Trial 2 Trial 3 P abs initial (kPa) 133. 79 131. 12 130. 66 141. 08 141. 23 141. 92 150. 67 150. 82 150. 51 161. 50 160. 84 161. 10 P abs inter (kPa) 119. 47 121. 58 123. 09 125. 20 130. 81 136. 96 139. 80 140. 50 133. 35 151. 66 151. 97 141. 70 P abs Final (kPa) 123. 30 123. 72 124. 83 129. 07 133. 17 137. 50 142. 03 142. 64 137. 44 150. 04 150. 14 145. 77 Table 1: Measured values from experiment 2 converted into the absolute pressure values P(abs). The formula used was P abs = P atm +P gauge 0.3 bars 0.4 bars 0.5 bars 0.6 bars rev= 𝘠 1.331 1.257 1.291 0.987 Table 2: Using the values from the figure 5, we calculated the mean specific heat capacity ratio for each pressure (we use the sum∑ of all Trials for each P).

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Figure 7: Pressure (yellow line) readings from Experiment 2 on a “P vs t” Graph Experiment 3 Sample Calculations Formulas used: P abs = P atm +P gauge where P atm is 102 kPa

and Figure 8:Equation 1.17 [1] P abs = 102 kPa + 29.27 kPa = 131.27 kPa 𝘠 irrev = ((131.27 kPa/102 kPa) - 1)/((131.27 kPa/108.40 kPa) - 1) = 1.36

Figure 9: Pressure (yellow line) readings from Experiment 3 on a “P vs t” Graph 0.3 bars Trial 1 Trial 2 Trial 3 P abs initial (kPa) 131.27 131.42 131.28 P abs inter (kPa) 102.76 102.73 102.75 P abs Final 108.40 108.40 108.41

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(kPa) Table 3: Measured values from experiment 3 converted into the absolute pressure values P(abs). The formula used was P abs = P atm +P gauge 0.3 bars Trial 1 Trial 2 Trial 3 𝘠 irrev = 1.360 1.358 1.361 Table 4: Specific Heat Capacity Ratio from each Trial on Table 3 DISCUSSION Experiment 1 During the brief depressurization of the pressure vessel, changes in both temperature and pressure were observed. The internal pressure of the vessel, initially stable at 0.3 bars (30kPa), dipped below the 0.25 bars (25kPa) and began augmenting until it stabilized at 0.25 bars (25kPa) (+/- 0.05 bars). A dip in the form of a reciprocal function was observed for the temperature line and a gradual increase was noted. The temperature stabilized once it matched that of the ambient temperature (23.4 ). ℃ Due to the atmospheric temperature of the room being higher than that of

the vessel after the brief depressurization, the increase of internal temperature of the vessel can be explained by the second law of thermodynamics which states that the entropy of an isolated system always increases. In other words, the system is attempting to reach equilibrium with that of the system surrounding it (we consider the entire lab room to be the latter). Because temperature and pressure are directly proportional, the increase of temperature causes the increase in pressure. From the specific heat capacity at constant pressure formula ( u = U / m ) where u is specific heat capacity at constant pressure, U is enthalpy and m is the mass of the system, we know that “heat energy [...] added at constant pressure [...] causes gas to expand, [thus] expanding the container” [1]. In this lab the container was rigid and could not expand, therefore no expansion of volume occurred (constant volume). We can also prove this using the first law of a closed system formula: dE / dt = Q − W + m 1 ( h 1 +( v 1 2 / 2 )+ g 1 z 1... )− m ( h 2 +( v 2 2 / 2 )+ g 2 z 2... ) dE/dt= change of internal energy over time Q= internal heat W= work m= mass of the system h= enthalpy v= velocity g= gravitational constant z= height For experiment 1 of this lab, all but Q and h cancel out which in turn leaves us with Q = h 2 − h 1 . This formula, for an ideal gas becomes Q = Cv ( T 2 − T 1 ) . Experiment 2 Calculating the average specific heat ratios yielded 1.331, 1.257, 1.291, and 0.987 for 30 KPa, 40 KPa, 50 KPa, and 60 KPa respectively. The given reference value for the specific heat ratio of air at room temperature was 1.401 [1]. Figure 8 shows the variation of the specific heat ratio given different ambient temperatures [1]:

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Figure 10: Relationship Between Specific Heat Ratio and Temperature of Ambient Air [1] The lowest deviation was found at 30 KPa with a 5.00% offset compared to 10.28% at 40 KPa, 7.85% at 50 KPa, and 29.55% at 60 KPa. This is to be expected as our point of reference remains the same while the system experiences changes in pressure which, from figure 4, we see is the driving variable for the specific heat capacity ratio. The offsets can also be explained by the increase in intermolecular activity as the pressure increases [1]. As the pressure increases, the compressibility factor will vary further from ideal gas behaviors and thus higher offsets are expected at higher pressures. Despite this, the offsets are minor and the pressures are still at low enough levels to consider the calculated values as valuable approximations.

In this lab, no heat is added or removed from the closed system, therefore this process is by definition adiabatic. We consider the pressure vessel (see figure 1 ) system to be isolated. When the ball valve goes from open to closed, the temperature goes down and the gas expands out of the vessel and into the lab room [1]. Because the system’s change is that of volume, we use the Work= ∫ PdV to demonstrate that Work = P (pressure) * (V2-V1). By definition, an adiabatic system is one which is isolated and in which energy is transferred through work [1]. No heat is applied or taken from the system, t herefore, the system is adiabatic. Experiment 3 The 𝘠 rev (mean specific heat capacity ratio) result obtained in Experiment 2 for the 0.3 bar trial using the formula from figure 7 was 1.331, whereas the result obtained from the experiment 3 using the formula from figure 11 was 1.360. There is a difference of 0.0310 units or 2.132% from the first to the second result for the specific heat capacity ratios. We can assume the reason for the lower results in experiment 2 is a bigger drop in temperature from the first to the second result which was to be expected as the specific heat capacity ratio is a property of the gas under specific conditions, in which these conditions were very similar. Experiment 2’s results had a 5.00% difference from the theoretical value of 1.401, whereas experiment 3 had a 2.93% difference. The offset between the two experiments is likely due the fluctuations in initial, intermediate, and final pressures as well as air leaks or issues with the sensors. The offset from the theoretical value is likely largely due to the differences in the ambient conditions used for the experiment versus the theoretical value. A reversible process is one during which the initial and final conditions of the system end up the same (volume, internal energy, etc.) all while not affecting the surrounding environment. To help us explain this, we considered the initial state values to have been 0 bars of pressure, ambient temperature and a rigid container (we do not have the dimensions of the Perfect Gas Expansion Apparatus but the rigidity of the vessel implies that initial volume= final volume). Furthermore, we considered the state during which the apparatus was filled with air at 0.3 bars to be the intermediary state . We considered the final state to be that of the vessel after having fully opened the ball valve and closed it again. The final state therefore contained the values of 0 bars of pressure, ambient temperature and the same volume as the initial volume ( see figure 6 ). We can also prove this using the first law of a closed system equation: dE / dt = Q − W + m 1 ( h 1 +( v 1 2 / 2 )+ g 1 z 1... )− m ( h 2 +( v 2 2 / 2 )+ g 2 z 2... ) dE/dt= change of internal energy over time Q= internal heat

W= work m= mass of the system h= enthalpy v= velocity g= gravitational constant z= height We cancel out the change of internal energy as the system is in a steady state; no work is done (volume remains constant); the mass of the system remains the same, and all other energies cancel out. This leaves us with the formula Q = h 2 − h 1 . This formula, for an ideal gas becomes Q = Cv ( T 2 − T 1 ) . In this case, because the initial temperatures are both the ambient temperature, Q= 0. This demonstrates equilibrium in the initial and final states. Although expansion of a gas is never truly reversible, due to adiabatic expansion against atmospheric pressure possibly causing a change in entropy, experiment 3 shows that the results of the irreversible process are are similar enough to the reversible process results that one can satisfactorily conclude that it is also a reversible process [1]. CONCLUSION Experiment 1: ● Brief depressurization of the pressure vessel causing changes in temperature and pressure (in vessel). ● The temperature stabilizes when it reaches the ambient temperature (23.4 ). This ℃ follows the second law of thermodynamics. ● In this experiment, the vessel was rigid, so there was no change in volume. ● Therefore, in the first law of thermodynamics, we are left with Q = Cv ( T 2 − T 1 ) .

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Experiment 2: ● From Figure 4, we see pressure is the driving variable for the specific heat capacity ratio. ● Offsets explained by increase in molecular activity when the pressure increases. ● Higher offsets are expected at higher pressures. ● Pressure vessel is isolated. ● The ball valve goes from open to close causing the temperature to go down and the gas to expand out of the vessel. ● There is a change in volume, so W= P (pressure) * (V2-V1). ● In this system, there is no heat change, it is therefore adiabatic. Experiment 3: . ● We assume the reason for lower results is a bigger drop in temperature. ● Experiment 2’s results had a 5.00% difference from 1.401. ● Experiment 3 had a 2.93% difference. ● The offset between the two is likely due to fluctuations in pressures and air leaks or issues with sensors. ● The offset from the theoretical value is likely due to differences in the ambient conditions used for the experiment vs the theoretical value. ● The formula becomes Q = Cv ( T 2 − T 1 ) but in this case, Q=0. This demonstrates equilibrium. in the initial and final states. ● Due to adiabatic expansion against atmospheric pressure, experiment 3 shows that the results of the irreversible process are are similar enough to the reversible process results that one can satisfactorily conclude that it is also a reversible process. SOURCES OF MEASUREMENT ERRORS One of the main sources of error could be attributed to the fluctuations of the pressure and temperature displayed by the hardware. Due to having a time limit, the values recorded may not have been fully stabilized before proceeding with further steps. To compound that, the ball valve may have been slowly releasing pressure due to imperfect sealing, thus not only making stabilizing difficult, but impossible. Even if no relative percentage error was mentioned on the machine used, it is known that for those types of experiment, there is always a relative error percentage. Since this was a modern lab with computer technology, it is assumed that the relative error percentage would be

small. Nevertheless, due to significant figures not always being carried over, the final results may have significant errors. Finally, it was not possible to get the same pressure values for the different trials, which may have led to results being a little less accurate. REFERENCES [1] Department of Mechanical, Industrial and Aerospace Engineering. THERMODYNAMICS II MECH 351 LABORATORY MANUAL (WINTER 2024). CONCORDIA UNIVERSITY; Engineering and Computer Science [2] K. M. Tenny and J. S. Cooper, “Ideal Gas Behavior,” in StatPearls , Treasure Island (FL): StatPearls Publishing, 2024. Accessed: Jan. 26, 2024. [Online]. Available: http://www.ncbi.nlm.nih.gov/books/NBK441936/

[3] “Specific Heats.” Accessed: Jan. 30, 2024. [Online]. Available: https://www.grc.nasa.gov/www/k-12/airplane/specheat.html

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