Answer to Question #246208 in Atomic and Nuclear Physics for Ejdjdhdh

Question

solve a uniformly capillary tube closed at one end contain dry air trapped by a thread of mecury 8.5*10m long. when the tube was held horizontally the length of the air column was 5.0^-2m when it was held vertically with the closed end downward length was 4.5*10^-2m determine the value of atmosphere pressure ( g=10m/s^-2) density of mecury 1.36*104kgm^-3

Answer 1

$P_1V_1=P_2V_2$ ..........(i)

$V_2=\frac{P_1V_1}{P_2}...........(ii)$

$P_1V_1=h_1(h_m+atm)$

$P_2=atm$

$V_2(horizntal)=5\times 10^{-2}m$

$h_1(closed)=4.5\times 10^{-2}m$

$h_m=8.5\times 10^{-2}m$

substituting these values in equation (ii) we have

$V_2=\frac{P_1V_1}{P_2}=\frac{h_1(h_m+atm)}{atm}$

$5\times 10^{-2}=\frac{4.5\times 10^{-2}(8.5\times 10^{-2} +atm)}{atm}$

$0.05atm=0.003825+0.045atm$

$0.005atm=0.003825$

$atm=0.765mHg=76.5cmHg$