An experimenter using a gas thermometer found the pressure...

An experimenter using a gas thermometer found the pressure at the triple point of water $\left(0.01^{\circ} \mathrm{C}\right)$ to be $4.80 \times 10^{4} \mathrm{Pa}$ and the pressure at the normal boiling point $\left(100^{\circ} \mathrm{C}\right)$ to be $6.50 \times 10^{4} \mathrm{Pa}$ , (a) Assuming that the pressure varies linearly with temperature, use these two data points to find the Celsius temperature at which the gas pressure would be zero (that is, find the Celsius temperature of absolute zero). (b) Does the gas in this thermometer obey Eq. $(17.4)$ precisely? If that equation were precisely obeyed and the pressure at $100^{\circ} \mathrm{C}$ were $6.50 \times 10^{4} \mathrm{Pa}$ , what pressure would the experimenter have measured at $0.01^{\circ} \mathrm{C} ?$ (As we will learn in Section $18.1, \mathrm{Eq} .(17.4)$ is accurate only for gases at very low density.)

Key Concepts

Gas Thermometry

This concept involves measuring temperature by exploiting the fact that, for a gas kept at constant volume, its pressure is directly related to its temperature. Gas thermometers use this predictable relationship, often employing a nearly ideal gas, to determine temperature by comparing measured pressures with known reference points.

Linear Extrapolation

Linear extrapolation is a mathematical technique used to extend a known linear relationship beyond the measured data points. In the context of gas thermometry, extrapolating the pressure–temperature line to the point where the pressure would theoretically be zero provides an estimate of absolute zero, a fundamental reference temperature.

Absolute Zero

Absolute zero is the theoretical temperature at which a system reaches its lowest possible energy state; in the ideal gas model, it is the temperature at which the pressure of a gas would vanish if the linear pressure–temperature relationship held down to that limit. Determining absolute zero through extrapolation from higher temperature data is a classical experiment in thermodynamics.

Ideal versus Non-Ideal Gas Behavior

The ideal gas law posits a direct proportionality between pressure and temperature at constant volume, which is essential for gas thermometry. However, in real gases, deviations from ideal behavior occur at higher densities or lower temperatures. Recognizing these discrepancies is important, as they can lead to slight differences between experimentally measured values and those predicted by the ideal gas law.

Transcript

00:01 So here would be the temperature pressure relation.

00:05 We can say tp relation.

00:07 And this is going to be equal to p sub 2 minus p sub 1.

00:12 Would be equal to gamma times t sub 2 minus t sub 1.

00:20 Or temperature sub 2 minus temperature sub 1.

00:23 So to find gamma, we would simply say p sub 2 minus p sub 1 and then divide by the change in temperature.

00:32 And this would be equal to 6 .5 .0 minus 4 .80 times 10 to the 4th pascals.

00:46 And then this would be divided by, of course, my apologies, 100 degrees celsius minus 0 degrees celsius.

01:00 And we find that gamma is going to be equal to 170 pascals per degree celsius.

01:08 So this would be the, that would be gamma.

01:14 And then if we wanted to find the true temperature of absolute zero, we can say that t sub 2 is going to be equal to t sub 1 plus p sub 2 minus p sub 1 divided by gamma.

01:29 Now at absolute zero, the temperature is going to be, sorry, the pressure is going to be 0 pascals.

01:36 So we can actually eliminate this term because the final pressure will be zero.

01:40 Pascal's and this was going to be equal to 0 .01 degrees celsius.

01:47 The pressure at 0 .01 degrees celsius is 4 .80 times 10 to the 4th pascals and then this would be divided by of course 170 pascals per degree celsius and we find that this is going to give us negative 282 degrees celsius so for this particular system this temperature negative 282 degrees celsius this would be absolute zero at absolute zero as a reminder all atom movement ceases so there is no there's actually no movement of the atoms within the solid so that's that's what defines absolute zero the pressure is going to be zero and then for part b, we're using equation 17 .4.

02:46 We know that this is only possible for low density gases.

02:51 So t sub 2 divided by t sub 1 would be equal to p sub 2 divided by p sub 1...

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