Question

An experimenter using a gas thermometer found the pressure at the triple point of water (0.01°C) to be $\mathbf{4}\mathbf{.}\mathbf{80}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{Pa}$and the pressure at the normal boiling point (100°C) to be$\mathbf{6}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{Pa}$. (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°C were$\mathbf{6}\mathbf{.}\mathbf{40}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{Pa}$, what pressure would the experimenter have measured at 0.01ºC?

Short answer

(a) The Celsius temperature at which the gas pressure is zero is $-282°C$.

(b) The pressure measured at 0.01ºC is $4.76\times {10}^4\mathrm{Pa}$which suggests that the equation is not precisely followed.

Step-by-step solution

(a) Determination of Celsius temperature at which the gas pressure is zero.

The linear relation of pressure and temperature is,

${\mathrm{p}}_2={\mathrm{p}}_1+\mathrm{y}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)$

Solve for $\mathrm{y}$,

$y=\frac{p_2-p_1}{T_2-T_1}$

Substitute the values given to obtain the value of $y$,

$$\begin{gathered} y=\frac{6.50\times {10}^4\mathrm{Pa}-4.80\times {10}^4\mathrm{Pa}}{100°\mathrm{C}-0.01°\mathrm{C}} \\ =170.0Pa/C° \end{gathered}$$

Now for zero gas pressure, p=0. Thus, The temperature at the zero pressure of the gas is,

$$\begin{gathered} T=T_1-\frac{p_1}{y} \\ 0.01°C-\frac{4.80\times {10}^{4}Pa}{170.0Pa/C°} \\ =-282°C \end{gathered}$$

(b) Determination of the pressure at the temperature 0.01ºC.

For this case,

$$\begin{gathered} {\mathrm{T}}_1=100°\mathrm{C}+273.15 \\ =373.15\mathrm{K} \end{gathered}$$ and $$\begin{gathered} {\mathrm{T}}_2=0.01°\mathrm{C}+273.15 \\ =273.16\mathrm{K} \end{gathered}$$

Use equation, $\frac{T_2}{T_1}=\frac{P_2}{P_1}$, Here, T is in units of K in order to obtain the value of pressure.

localid="1664357082266" $$\begin{gathered} P_2=P_1\left(\frac{T_2}{T_1}\right) \\ {\mathrm{P}}_2=6.50\times {10}^4\mathrm{Pa}\left(\frac{273.16\mathrm{K}}{373.15\mathrm{K}}\right) \\ =4.76\times {10}^4\mathrm{Pa} \end{gathered}$$

The measured pressure is $4.80\times {10}^4\mathrm{Pa}$and the calculated pressure is $4.76\times {10}^4\mathrm{Pa}$. Therefore, the equation is not precisely followed.