Measuring Pressure — Question 5

Question 5

A barometer headspace need not be a perfect vacuum

A mercury barometer has column height $h=0.740\,\mathrm{m}$ above its reservoir surface. The sealed space above the mercury has known vapor pressure $p_v=0.20\,\mathrm{kPa}$. Take $\rho=13600\,\mathrm{kg/m^3}$ and $g=9.80\,\mathrm{m/s^2}$. Ignore capillarity.

1. Find atmospheric pressure and the error made by assuming a perfect vacuum above the column.

2. If vapor pressure rises by $0.30\,\mathrm{kPa}$ at fixed atmospheric pressure and fixed liquid density, by how much does the column height change?

Question 5 - Solution

A barometer headspace need not be a perfect vacuum

(a) Derivation and calculation

$$\boxed{p_{\mathrm{atm}}=p_v+\rho gh}.$$$$\rho gh=(13600)(9.80)(.740)=98627.2\,\mathrm{Pa}.$$$$\boxed{p_{\mathrm{atm}}=200+98627.2=98827.2\,\mathrm{Pa}=98.8272\,\mathrm{kPa}}.$$The perfect-vacuum assumption underestimates pressure by $.200\,\mathrm{kPa}$.

(b) Result and interpretation

With atmosphere and density fixed,$$0=\Delta p_v+\rho g\Delta h.$$$$\boxed{\Delta h=-\frac{300}{13600(9.80)}=-.00225090\,\mathrm m=-2.251\,\mathrm{mm}}.$$The column falls as the headspace pressure rises.

Physical check

Pressure differences, rather than tube length or an assumed vacuum, determine the supported vertical head.