Solved Problem on Temperature
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In a constant-pressure gas thermometer, the physical quantity is the gas volume. The calibration of the thermometer is given by the graph shown in the figure. Determine:
a) The equation of that thermometer;
b) When the volume of the gas is 130 cm3, what will be the temperature of the gas?



Problem data:
  • For volume:
    • First calibration point:    V1 = 100 cm3;
    • Second calibration point:    V2 = 200 cm3;
  • For temperature:
    • First calibration point:    t1 = 0 °C;
    • Second calibration point:    t2 = 273 °C.
Problem diagram:

We chose any point (V, t) of the graph, two other points are (V1, t1)=(100, 0) and (V2, t2)=(200, 273).
Figure 1

Solution

a) To find the equation of this thermometer, we have
\[ \bbox[#99CCFF,10px] {\frac{g-g_{1}}{g_{2}-g_{1}}=\frac{t-t_{1}}{t_{2}-t_{1}}} \]
where physical quantity g for this problem is the volume of the gas V
\[ \frac{V-V_{1}}{V_{2}-V_{1}}=\frac{t-t_{1}}{t_{2}-t_{1}} \]
substituting the problem data, we have
\[ \begin{gather} \frac{V-100}{200-100}=\frac{t-0}{273-0}\\ \frac{V-100}{100}=\frac{t}{273}\\ 273\times (V-100)=100t\\ 273V-273\times 100=100t\\ 100t=273V-27300\\ t=\frac{273}{100}V-\frac{27300}{100} \end{gather} \]
the equation that gives us the temperature as a function of volume t = f(V), will be
\[ \bbox[#FFCCCC,10px] {t=2.73V-273} \]

b) For V = 130 cm3, we have substituted this value in the equation found in the previous item to obtain
\[ t=2.73\times 130-273 \]
\[ \bbox[#FFCCCC,10px] {t=81.9 °\text{C}} \]
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