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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?

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Answer:

[tex]833.72\ \text{torr}[/tex]

Explanation:

[tex]P_g[/tex] = Pressure of gas = [tex]106\ \text{mmHg}[/tex]

[tex]P_a[/tex] = Atmospheric pressure = 0.958 atm

Absolute pressure is given by

[tex]P=P_g+P_a[/tex]

[tex]\Rightarrow P=\dfrac{106}{760}+0.958[/tex]

[tex]\Rightarrow P=1.097\ \text{atm}=1.097\times 760=833.72\ \text{torr}[/tex]

The pressure of the gas in the tank is [tex]833.72\ \text{torr}[/tex]

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