Question

The mercury level in the open arm of an open-end manometer is 106 mm above the level in the arm connected to a tank of gas. If the barometric pressure is 0.958 atm, what is the pressure (in torr) of the gas in the tank?

Answer 1

Answer:

$833.72\ \text{torr}$

Explanation:

$P_g$ = Pressure of gas = $106\ \text{mmHg}$

$P_a$ = Atmospheric pressure = 0.958 atm

Absolute pressure is given by

$P=P_g+P_a$

$\Rightarrow P=\dfrac{106}{760}+0.958$

$\Rightarrow P=1.097\ \text{atm}=1.097\times 760=833.72\ \text{torr}$

The pressure of the gas in the tank is $833.72\ \text{torr}$