Question

The mercury manometer shown in FIGURE P18.56 is attached

to a gas cell. The mercury height h is 120 mm when the cell is

placed in an ice-water mixture. The mercury height drops to

30 mm when the device is carried into an industrial freezer.

What is the freezer temperature?

Hint: The right tube of the manometer is much narrower than the

left tube. What reasonable assumption can you make about the gas

volume?

Short answer

$T_2=-27.9^{0}C$represents the volume of water.

Step-by-step solution

Given information

The mercury was at its highest point at the time. $h_1=120\mathrm{mm}$

The mercury reached its highest point at this point.$h_2=30\mathrm{mm}$

Initial temperature $T_1=273K$

The manometer's right-side tube is quite small compared to the opposite side, so the volume change is minimal and the volume remains constant. By substituting them, the ideal gas equation can be used to find the values of pressure and temperature.

formula used:

Ideal gas law

$PV=nRT$

Absolute pressure

$P=P_a+\rho gh$

${\mathrm{P}}_{\mathrm{a}}$-Atmospheric pressure

calculation

Description	Steps
Step:1 calculate the absolute pressure values	Initial pressure $$\begin{gathered} P_1=P_a+\rho gh \\ =101300Pa+\left(13600\mathrm{kg}/{\mathrm{m}}^3\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 120\times {10}^{-3}\mathrm{m}\right) \\ =101300+15993.6 \\ {P}_1=117293\mathrm{Pa} \end{gathered}$$ Final pressure $$\begin{gathered} P_2=P_a+\rho g{h}_2 \\ =101300Pa+\left(13600\mathrm{kg}/{\mathrm{m}}^3\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 30\times {10}^{-3}\mathrm{m}\right) \\ =101300+3998 \\ {P}_2=105298\mathrm{Pa} \end{gathered}$$
Step:2 calculating temperature of the freezer	$$\begin{gathered} \frac{P_1}{P_2}=\frac{T_1}{T_2} \\ {T}_2=273K\left(\frac{105298P_a}{117293P_a}\right) \\ =245.1K \\ {T}_2=-27.9^0\mathrm{C} \end{gathered}$$

Therefore ,Temperature of the freezer is $T_2=-27.9^0\mathrm{C}$