IChO 1976 · Theory problem 6

Gas thermometer

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Temperature in a larger room should be measured by means of a gaseous thermometer. A glass tube with the internal volume of 80 cm³ was for this purpose filled with nitrogen at a temperature of 20 °C and a pressure of 101.325 kPa. The tube was then slowly and steadily moved throughout the room. Due to the thermal expansion the gas at the higher temperature escapes from the tube and is captured above the liquid whose vapour pressure is negligible. The total volume of the gas escaped from the tube was 35 cm³ at a temperature of 20 °C and a pressure of 101.325 kPa.

Problems:

6.1 How many moles of nitrogen were used to fill the glass tube?

6.2 How many moles of nitrogen escaped from the tube at the higher temperature?

6.3 Calculate the average temperature in the room under investigation if the thermal expansion of the glass tube is assumed to be negligible.

6.4 Does anything change if instead of pure nitrogen, a mixture containing 50 volume % of nitrogen and 50 volume % of hydrogen is used?

Official English text of the problem.

Answers

  • 6.1 Initially, \(3.33\times10^{-3}\ \mathrm{mol}\) of nitrogen, \(\ce{N2}\), N#N.
  • 6.2 Escaped nitrogen: \(1.46\times10^{-3}\ \mathrm{mol}\).
  • 6.3 The inferred room temperature is approximately \(\boxed{521\ \mathrm{K}=248\ ^\circ\mathrm{C}}\).
  • 6.4 The inferred temperature and total mole amounts are unchanged for an ideal, nonreacting mixture. For a mixture that remains well mixed, nitrogen and hydrogen, \(\ce{H2}\), [H][H], each contribute \(1.66\times10^{-3}\ \mathrm{mol}\) initially and \(7.28\times10^{-4}\ \mathrm{mol}\) to the escaped gas.

Solution

Treat the gases as ideal and take their pressure during expansion to remain equal to the stated atmospheric pressure. Use

\[ pV=nRT, \qquad R=8.314\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}, \]

with the reference temperature

\[ T_0=20+273.15=293.15\ \mathrm{K}. \]

6.1

The initial volume of nitrogen is

\[ V_0=80\ \mathrm{cm^3}=0.080\ \mathrm{L}. \]

Therefore,

\[ n_0=\frac{pV_0}{RT_0} =\frac{(101.325\ \mathrm{kPa})(0.080\ \mathrm{L})} {(8.314\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}})(293.15\ \mathrm{K})} =3.3259\times10^{-3}\ \mathrm{mol}. \]

Thus,

\[ \boxed{n_0\approx3.33\times10^{-3}\ \mathrm{mol}}. \]

6.2

The escaped gas is measured at the same reference temperature and pressure. Since the collecting liquid has negligible vapour pressure, no vapour-pressure correction is required.

\[ n_{\mathrm{escaped}} =\frac{pV_{\mathrm{escaped}}}{RT_0} =\frac{(101.325\ \mathrm{kPa})(0.035\ \mathrm{L})} {(8.314\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}})(293.15\ \mathrm{K})} =1.4551\times10^{-3}\ \mathrm{mol}. \]

Hence,

\[ \boxed{n_{\mathrm{escaped}}\approx1.46\times10^{-3}\ \mathrm{mol}}. \]

6.3

The amount remaining in the tube is

\[ n_{\mathrm{remaining}} =n_0-n_{\mathrm{escaped}} =1.8708\times10^{-3}\ \mathrm{mol}. \]

Because both measured volumes refer to the same temperature and pressure,

\[ \frac{n_{\mathrm{remaining}}}{n_0} =\frac{80-35}{80} =\frac{45}{80}. \]

The glass tube retains its volume of \(80\ \mathrm{cm^3}\). Applying the ideal gas law to its initial and heated states gives

\[ pV_0=n_0RT_0=n_{\mathrm{remaining}}RT. \]

Consequently,

\[ T=T_0\frac{n_0}{n_{\mathrm{remaining}}} =293.15\ \mathrm{K}\times\frac{80}{45} =521.16\ \mathrm{K}. \]

Converting to Celsius,

\[ t=(521.16-273.15)\ ^\circ\mathrm{C} =248.01\ ^\circ\mathrm{C}. \]

Thus the intended thermometer result is

\[ \boxed{T\approx521\ \mathrm{K},\qquad t\approx248\ ^\circ\mathrm{C}}. \]

6.4

For an ideal mixture of nitrogen, \(\ce{N2}\), N#N, and hydrogen, \(\ce{H2}\), [H][H],

\[ pV=\left(n_{\ce{N2}}+n_{\ce{H2}}\right)RT. \]

At a common temperature and pressure, volume fractions equal mole fractions. The specified mixture therefore has

\[ x_{\ce{N2}}=x_{\ce{H2}}=0.50. \]

The ideal gas law depends on the total amount of gas, irrespective of its molecular identity. Thus the initial total amount, escaped total amount, and inferred temperature remain unchanged.

The amount of nitrogen itself is halved. Assuming the escaping gas has the same composition as the well-mixed gas in the tube, each constituent has

\[ n_{i,0}=\frac{n_0}{2} \approx1.66\times10^{-3}\ \mathrm{mol}, \]

and

\[ n_{i,\mathrm{escaped}} =\frac{n_{\mathrm{escaped}}}{2} \approx7.28\times10^{-4}\ \mathrm{mol}. \]

The solution was checked against the official solution.