Measuring Pressure — Question 10

Question 10

A trapped gas makes a closed-end reading nonlinear

A vertical U-tube of uniform area contains water. Initially its two surfaces are level; the left side is open to atmosphere at $p_0=100\,\mathrm{kPa}$ and the right side traps gas of length $L_0=0.20\,\mathrm{m}$ at $p_0$. Gas of higher pressure is then applied on the left. The right water surface rises by $x=0.020\,\mathrm{m}$ while the left falls equally. The trapped gas obeys the supplied isothermal law $pV=\text{constant}$. Use $\rho=1000\,\mathrm{kg/m^3}$ and $g=10\,\mathrm{m/s^2}$.

1. Find the final trapped-gas pressure and applied left pressure.

2. Explain why treating the closed end as a vacuum is invalid, and identify the singular limit of the ideal model.

Question 10 - Solution

A trapped gas makes a closed-end reading nonlinear

(a) Derivation and calculation

$$V_0=AL_0,\quad V_R=A(L_0-x),\quad p_RV_R=p_0V_0.$$$$\boxed{p_R=\frac{p_0L_0}{L_0-x}=\frac{100(.200)}{.180}=111.111\,\mathrm{kPa}}.$$$$h=2x=.0400\,\mathrm m,\quad\boxed{p_L=p_R+2\rho gx=111.511\,\mathrm{kPa}}.$$

(b) Result and interpretation

$$p_L(x)=\frac{p_0L_0}{L_0-x}+2\rho gx,\qquad0\le x<L_0.$$$$x\to L_0^-\Rightarrow V_R\to0^+,\quad p_R\to\infty.$$The gas is neither a vacuum nor a fixed-pressure reference. The ideal law has a zero-volume singularity and cannot be extended beyond it.

Physical check

At $x=0$, both pressures return to $p_0$, recovering the initial level-surface condition.