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vaieri [72.5K]
3 years ago
5

Why is it better to conduct an experiment more than once?

Chemistry
1 answer:
lilavasa [31]3 years ago
8 0

The first reason to repeat experiments is simply to verify results. Different science disciplines have different criteria for determining what good results are. Biological assays, for example must be done in at least triplicate to generate acceptable data. Science is built on the assumption that published experimental protocols are repeatable.


2)      The next reason to repeat experiments is to develop skills necessary to extend established methods and develop new experiments. “Practice make perfect” is true for the concert hall and the chemical laboratory.


3)      Refining experimental observations is another reason to repeat. Maybe you did not follow the progress of the reaction like you should have.


4)      Another reason to repeat experiments is to study and/or improve them in way. In the synthetic chemistry laboratory, for example, there is always a desire to improve the yield of a synthetic step. Will certain changes in the experimental conditions lead to a better yield? The only way to find out is to try it! The scientific method informs us that it is best to only make one change at a time.


5)  The final reason to repeat an extraction, chromatographic or synthetic protocol is to produce more of your target substance. This is sometimes referred to scale-up.

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A compression, at a constant pressure of 140 kPa, is performed on 4.0 moles of an ideal monatomic gas (Cv = 3/2 R). The compress
kobusy [5.1K]

<u>Answer:</u> The change in internal energy of the gas is 29.414 kJ.

<u>Explanation:</u>

To calculate the temperature of the gas at different volumes, we use ideal gas equation:

PV=nRT

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We are given:

Conversion used:  1m^3=1000L

P=140kPa\\V=0.23m^3=260L\\n=4.0mol\\R=8.31\text{L kPa }mol^{-1}K^{-1}

Putting values in above equation:

140kPa\times 260L=4mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times T_i\\\\T_i=1095.06K

  • When volume = 0.12m^3

We are given:

P=140kPa\\V=0.12m^3=120L\\n=4.0mol\\R=8.31\text{L kPa }mol^{-1}K^{-1}

Putting values in above equation:

140kPa\times 120L=4mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times T_f\\\\T_f=505.41K

  • To calculate the change in internal energy, we use the equation:

\Delta U=nC_v\Delta T=nC_v(T_f-T_i)

where,

\Delta U = change in internal energy = ?

n = number of moles = 4.0 mol

C_v = heat capacity at constant volume = \frac{3}{2}R

T_f = final temperature = 1095.06 K

T_i = initial temperature = 505.41 K

Putting values in above equation, we get:

\Delta U=4\times \frac{3}{2}\times 8.314J/K.mol\times (505.41-1095.06)\\\\\Delta U=29414.1J

Converting this value in kilojoules, we use the conversion factor:

1 kJ = 1000 J

So, 29414.1J=\frac{1kJ}{1000J}\times 29414.1J=29.414kJ

Hence, the change in internal energy of the gas is 29.414 kJ.

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