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June 5th, 2016, 05:06 PM   #1
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centigrade scale, Celsius scale and thermodynamic scale

Hi can someone give me some info on what this question is asking please.

A constant gas thermometer and a thermocouple are both used to measure the boiling point of a certain liquid. The readings, together with readings for ice and steam for water are

Gas thermometer pressure (Kpa) Thermocouple voltage (mV)
Ice point 101 0
steam point 138 5.4
liquid's boiling point 124 3.4

Calculate the liquid's boiling point on the Celsius scale and on the centigrade scale; comment on your results. What is the boiling point of the liquid on the thermodynamic scale?

So as I understand there is not much difference between centigrade scale and Celsius scale. So I would approach this question with the formula

$\displaystyle \theta=\frac{x_\theta-x_0}{x_100-x_0}*100$

so

$\displaystyle X_100=\text{steam point}$
$\displaystyle X_\theta = \text{liquid's boiling point}$
$\displaystyle X_0 = \text{ice point}$

sub in the values and compare the results and I will assume that the gas thermometer is more accurate compared to the thermal couple due to heat loss to the surrounding for example.

But I don't understand what it means by thermodynamic scale. In my book, it states that the thermodynamic scale is:

$\displaystyle T=\frac{pV}{pV_tr}*273.16$ how would I use this to calculate the value they want???

Big thanks in advance.

Last edited by skipjack; June 5th, 2016 at 06:42 PM.
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June 6th, 2016, 03:27 AM   #2
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Centigrade is just an old fashioned name for the same scale as Celsius. There is no difference between the two.

To convert the pressure reading into temperature, use

$\displaystyle pV = nRT$

where $\displaystyle T$ is in Kelvin. Because you have no fiducial value, you need to calculate the values relative to one particular state. I recommend calculating them relative to the "ice point", because we know that 0 degrees Celsius is defined to be the temperature at the ice point (actually, it's the position of the triple point of water, but whatever).

To convert the thermocouple voltage to temperature requires knowledge of the particular thermocouple you are using. In the absence of that, you could perhaps assume that the thermocouple voltage is linear with temperature and then use linear interpolation to estimate the value.
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June 6th, 2016, 03:52 AM   #3
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I did eventually calculate this morning, which i pv=nrt. Thanks for the response and the advice much appreciated
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