Question

A barometer reads $75\;cm$ of Hg. When $2\;cc$ of air at atmospheric pressure is introduced into the space above the mercury level, the volume of this space becomes $50\;cc$. The length by which the mercury column descends is:
A.$3\;cm$
B.$6\;cm$
C.$1.5\;cm$
D.$25\;cm$

Answer 1

Hint: From the question, we understand that the expansion of volume of air in the mercury column occurs under isothermal conditions. This means that the product of the pressure exerted and volume remains constant for any process in the system. Use this to equate the initial and final conditions of pressure and volume and consequently determine the final height of the mercury column, the difference of which will give the length by which the mercury column descends.

Formula Used:
Barometric formula: $P = \rho gh$
Boyle’s Law: $P_2 V_2 = P_1 V_1$

Complete answer:
We know that a barometer is a device that is used to measure localized atmospheric pressure. It works on a pretty simple principle wherein, an inverted glass tube stands in a bath of mercury, and air pressure pushes down on the surface of the mercury making some of the mercury rise up the tube. The mercury rises higher for greater pressures.
For a height h of mercury in the tube, the barometer is calibrated in accordance with the following expression, relating the height of mercury in the tube to the pressure exerted:
$P = \rho g h$, where $\rho$ is the density of the barometric fluid (here mercury), g is the acceleration due to gravity, and h is the height of the mercury column in the glass tube.
Now, when a volume of air is introduced into the space above the mercury level in the inverted glass tube, the pressure pushes down on the mercury and it descends down.
Therefore, we have, initially,
$P_1 = \rho g (75)$ and $V_1 = 50-2 = 48\;cc$
And finally,
$P_2 = \rho g (h)$ and $V_2 = 50\;cc$
This decrease (increase) in the volume of mercury (air) in the tube with an increase in pressure occurs as an isothermal process where the barometer sees no change in the overall temperature.
This means that the product of the pressure and volume are observed to be constant. This is given by Boyle’s law: $PV = constant$
$\Rightarrow P_1 V_1 = P_2 V_2 \Rightarrow P_2 = \dfrac{P_1 V_1}{V_2}$
$\Rightarrow \rho g h = \dfrac{\rho g (75) \times 48}{50} \Rightarrow h = \dfrac{75 \times 48}{50} = 72\;cm$
This means that the length of the mercury column decreases by $75 – 72 = 3\;cm$

Therefore, the correct option is A. $3\;cm$

Note:
Remember that we are asked to find the change in the height of the mercury column and not the overall height that the mercury column obtains as a result of the isothermal expansion of air. Therefore, always be cautious of what parameter we are asked to determine.
Also, remember that the barometric pressure measured by our device is nothing but the hydrostatic pressure or a compressive stress that is caused by the weight of air above the mercury surface. This is also how we justify the decreasing atmospheric pressure with increasing altitude, since there is less overlying mass with increasing elevation.