Question

If the gas inside the flask in the above exercise is cooled so that its pressure is reduced to a value of 715.7 torr, what will be the height of the mercury in the open ended arm? (Hint: The sum of the heights in both arms must remain constant regardless of the change in pressure.) (a) 49.0 mm, (b) 95.6 mm, (c) 144.6 mm, (d) 120.1 mm.

Answer 1

Answer:

(a) 49.0 mm

Explanation:

#First we need to calculate the height of the mercury arm:

Reduced pressure is 715.7torr, Initial Pressure is 797.3torr, Required pressure is:

$P_g_a_s=P_a_t_m+P_h$  

The difference in the two arms will give the pressure difference between the gas placed in the flask attached to an open-end mercury manometer.

$P_g_a_s=P_a_t_m+P_h\\797.3=P_a_t_m+(136.4mm-103.8mm)\\P_a_t_m=764.7mm\\\\h_t_o_t_a_l=136.4mm+103.8mm=240.2mm\\\\P_g_a_s=P_a_t_m+\bigtriangleup h\\\bigtriangleup h=P_g_a_s-P_a_t_m\\\\\bigtriangleup h=715.7mm-764.7mm\\=-49mm$

Hence the height of the mercury in the arm is 49.00mm