Question

A volume of $190.0$ ml of ${N_2}$ was collected in a jar over water at some temperature, water level inside and outside the jar standing at the same height. If a barometer reads 740 mm Hg and aqueous tension at the same temperature of the experiment is 20 mm Hg, the volume of the gas at 1 atm pressure and at the same temperature would be?
A.$185.0$ ml
B.$180.0$ ml
C.$195.0$ ml
D.$200$ ml

Answer 1

We know that if the initial pressure and volume of fixed moles of nitrogen gas at constant temperature are ${{\text{P}}_{{{\text{N}}_{\text{2}}}}}$ and ${{\text{V}}_{{{\text{N}}_{\text{2}}}}}$ , and after the expansion or compression the total pressure and total volume of the gas occupied will be ${{\text{P}}_{\text{T}}}$ and ${{\text{V}}_{\text{T}}}$ respectively, then according to the Boyles law:

Formula used: ${{\text{P}}_{{{\text{N}}_{\text{2}}}}}{{\text{V}}_{{{\text{N}}_{\text{2}}}}}{\text{ = }}{{\text{P}}_{\text{T}}}{{\text{V}}_{\text{T}}}$ (equation 1)

Complete step by step answer
We are given the volume of nitrogen over water as $190.0$ mL
${{\text{V}}_{{{\text{N}}_{\text{2}}}}} = 190.0{\text{mL}}$
We know that the pressure read by the barometer is 740 mm Hg
So the total pressure becomes:
${{\text{P}}_{\text{T}}} = 740{\text{mmHg}}$
And the aqueous tension is given as 20 mm Hg
Now, we know that the total pressure is equal to the pressure exerted by the gas plus the aqueous tension (AT).
${{\text{P}}_{\text{T}}} = {{\text{P}}_{{{\text{N}}_{\text{2}}}}} + {\text{AT}}$
Now by substituting the value:
$740{\text{mmHg}} = {{\text{P}}_{{{\text{N}}_{\text{2}}}}} + 20{\text{mmHg}}$
By taking all the numerical value on one side we get:
$740{\text{mmHg}} - 20{\text{mmHg}} = {{\text{P}}_{{{\text{N}}_{\text{2}}}}}$
We get the pressure exerted by the nitrogen gas as:
${{\text{P}}_{{{\text{N}}_{\text{2}}}}} = 720{\text{mmHg}}$
Now, we are asked the volume of nitrogen gas at 1 atm pressure.
So, the total pressure now becomes 1 atm and we know that 1 atm is equal to 760 mm Hg.
${{\text{P}}_{\text{T}}} = 1{\text{atm}} = 760{\text{mmHg}}$
Now substituting the value in equation 1 we get:
$720{\text{mmHg}} \times 190{\text{mL}} = 760{\text{mmHg}} \times {{\text{V}}_{\text{T}}}$
By taking all the numerical value on one side we get:
$\dfrac{{720{\text{mmHg}} \times 190{\text{mL}}}}{{760{\text{mmHg}}}} = {{\text{V}}_{\text{T}}}$
Now by solving we get the total volume as:
${{\text{V}}_{\text{T}}} = 180{\text{mL}}$

Therefore, we can conclude that the correct answer to this question is option B

Note: We know that to find the pressure of nitrogen gas we must cancel the aqueous tension from the barometric pressure. We must focus on unit conversion and convert the unit from atm to mm Hg as $1{\text{atm}} = 760{\text{mmHg}}$ .