Experiment to Study the Different Gas Laws

Aim

Aim of this experiment is to study the different gas laws

Objectives

1. Study and verify the Boyle’s and Gay Lussac gas laws associated with ideal gasses.

2. Study the isothermal and isochoric process

Theory

According to the definition a perfect ideal gas is one which does not have any volume, whose particles does not attack or repeal each other and there collision of these gasses is perfectly elastic. Ideal gasses are all those gasses that follow the gas laws in perfect manner.

Isothermal process

Isothermal process is one which follows the Boyle’s law and in this process gas is compressed or expanded at a constant temperature. For any ideal gas the produce of PV should be kept constant.

PV=C

Isochoric process

Isochoric process is one in which follows the Gay Lussac law and in this process gas is heated and cooled at a constant volume. For any perfect gas the ratio of pressure with temperature should remain constant at all stages. 

P/T=C

Isobaric Process

Isobaric process is one which follows the Charles law and in this process gas is heat at constant pressure. A perfect gas should have a constant ratio of volume with temperature explained by following equation

V/T=C

General Gas Equation

General gas equation is the combination of all three laws (Boyle’s Law, Charles Law and Gay Lussac Law) of gas and perfectly ideal gasses follow this law in perfect manner.

PV/T=C

General Gas equation

General gas equation is an equation obtain as a result of combination and manipulation of different gas laws and it can be written as follow

PV=mRT

Where 

P is pressure of gas

V is volume of chamber

m is mass of gas

R is gas constant

T is temperature of gas

Experimental 1: Isothermal Process

Procedure

·         First step was to find the temperature, pressure and volume of gas at ambient condition and note it in separate temperature

·         Second step was to open the discharge valve and set 5/2 way to A for compression and for expansion set it to B

·         Compressor was turn on to let the oil enter in compression condition and exist in expansion condition

·         Turn off the discharge valve and compressor when oil level reach the required height

·         Once the digital displace was stable temperature, pressure and volume of gas was recorded

·         Above step was repeated for different oil levels and data was recorded in table

Results

Compression

Table 1 ambient condition

	1	2	3
Temp	20.7	21	21.5
Pressure	0.99	1	1
Volume	2.96	2.96	2.96
PV	2.9304	2.96	2.96

Table 2 second case

	1	2	3	Average
Temp	21.4	21.8	22.2
Pressure	1.19	1.2	1.22	1.203
Volume	2.49	2.51	2.42	2.453
PV	2.9631	2.94	2.9524

Table 3 third case

	1	2	3	Average
Temp	21.1	21.4	21.7
Pressure	1.93	1.97	1.95	1.95
Volume	1.50	1.49	1.49	1.49
PV	2.895	2.9353	2.9055

Expansion

Table 4 first case

	1	2	3
Temp	19.9	20.6	20.8
Pressure	0.96	0.96	0.96
Volume	1.57	1.61	1.60
PV	0.048241206	0.046601942	0.046153846

Table 5 second case

	1	2	3
Temp	20	20.4	20.8
Pressure	0.61	0.62	0.62
Volume	2.54	2.51	2.55
PV	0.0305	0.030392157	0.029807692

Questions/Answers

a)      According to the results air is not according to the ideal gas in the case of compression and expansion. Reason is that the gas molecule does have some volume and they do apply some forces on other molecules. These forces are very small and if that much precision is not required it can be said that the air is an ideal gas because its produce of pressure and gas is almost constant

b)      No the results of produce of pressure and volume in table 2 and 3 is not constant and the man reason for this is that air is not a perfect gas and it pressure and volume does not change as it does in the case of perfect ideal gas

c)      No the result of produce of pressure and volume in table 4 and 5 is not constant and the reason is the same as it is for table 2 and 3

d)     R = 8314 J/Kmol K

m=PV/RT

m=(119500*0.002473)/(8314*294.8)

m=0.003493 Kg

m=PV/RT

m=(785666*0.00149)/(8314*294.4)

m=0.013855 Kg

Mass of gas in this case is more than that of the previous case and the reason of that is the high pressure and lower volume of oil in the container as compared to previous case and air is not an ideal gas due to which is get compressed and have more mass in high pressure and low volume condition.

Experiment 2 Isochoric Heating

Procedure

First of all discharge valve was closed

Then heater was turn on to increase the temperature from ambient to 35 C

Record pressure reading for different temperature

Results

Heating

Table 6 heating

Time in minute	Temperature in C	Pressure	P/T=C Pa/K
1	22.9	1.033	349.1044272
2	24.6	1.048	352.1505376
3	27.7	1.065	354.1735949
4	31.1	1.079	354.8174942
5	34.5	1.092	355.1219512

Cooling

Table 7 cooling

Time sec	Temperature in C	Pressure bar	P/T = C  Pa/K
1	37.9	1.012	325.5065938
2	37.4	1.009	325.064433
3	37	1.006	324.516129
4	36.4	1.004	324.4990304
5	35.9	1.002	324.376821
6	35.4	0.09	29.18287938
7	35	0.997	323.7012987
8	34.5	0.995	323.5772358
9	34	0.994	323.7785016
10	33.6	0.992	323.5485975

Questions

    In table 6 and 7 repetitions of the value in P/T column can be seen and this indicates that air follow isochoric process to some extend but there is some change in values which show air is not 100% ideal gas. There are three big reasons for that, one air molecules have some physical mass, second is that air molecules does attract or repeal each other and third is that the collision between air molecules is not 100 % elastic in nature.

From of P/T and time it can be said that heating process is more effective than cooling as it show sharper curve than cooling process. But it clearing show that air is not an ideal gas as its P/T graphs verses time is not a straight horizontal line as in the case ideal gas. During the cooling process of air the curve obtain at the P/T and time graphs show very little deflection from horizontal axis with respect to the heating curve. This mean that during cooling process at constant volume air is somewhat behaving like an ideal gas. This can be explain by the concept that during the heating process the collision of air molecules increase because of the extra energy provided to them. As the collision of air molecules is not perfectly elastic so due to this air deviate more form ideal behaviors but during cooling this process work in reverse direction. With cooling collision became less which allow air to behave more like ideal gas.

Conclusion

Aim of this experiment was to investigate the behaviors of air in isothermal and isochoric process. Isothermal process follows the Boyle’s law and isochoric process follows Gay Lussac law.  From the result that this experiment has shown it can be said that air as a gas is not an ideal gas because when it was tested as constant temperature to get PV = C at every point air show different produce of pressure and gas. When air was tested at constant volume to check the isochoric process then the P/T ratio of air was not constant at every point, although there was some repetition of value but it is not sufficient to called air an ideal gas. So based on results it can be said that air does not follow isothermal and isochoric process which are governed Boyle’s and Gay Lussac Law respectively.