Answer to Question #77736 in Molecular Physics | Thermodynamics for Rustam

Question

A 1 meter tube, closed at one end, is immersed in water completely. What is the open air pressure if the water rises 9 cm in the tube? Assume temperature remains constant.

Answer 1

According to the Boyle’s Law
p_1 V_1=p_2 V_2
We have
p_a V_1=pV_2 → p=(p_a V_1)/V_2 =p_a (S∙h)/(S∙(h-∆h) )=p_a h/(h-∆h)=101325∙1/(1-0.09)=111346 Pa