Question

Two gas bulbs A and B are connected by a tube having a stopcock. Bulb A has a volume of 100 mL and contains hydrogen. After opening the gas from A to the evacuated bulb B, the pressure falls down by 40%. The volume (mL) of B must be:
A. 75
B. 150
C. 125
D. 200

Answer 1

Hint – Let the initial pressure in bulb A be P. Then, pressure falls down by 40%, so final pressure will ${P_f} = \dfrac{{40}}{{100}}P$ and then assume the initial volume be V (volume of bulb B) and then final volume is (100 + V), then use Boyle's Law to solve.

Formula used - ${P_i}{V_i} = {P_f}{V_f}$

Complete answer:
We have been given that two gas bulbs A and B are connected by a tube having a stopcock.
Volume of bulb A = 100 mL
Also, given that pressure falls down by 40% after opening.
So, let us assume the initial pressure in bulb A be P and initial volume be V, i.e. volume of bulb B.
Now, as the pressure falls down to 40 % of the initial pressure, so we can write-
${P_f} = \dfrac{{40}}{{100}}P$
Also, the final volume will be (100 + V) mL
Now, using the Boyle’s Law-
${P_i}{V_i} = {P_f}{V_f}$
Here, Pi is the initial pressure, Pf is the final pressure, Vi is the initial volume and Vf is the final volume.
So, putting the values in the formula, we get-
$P \times 100 = \dfrac{{40}}{{100}}P \times (100 + V) \\\ \Rightarrow 100 = \dfrac{2}{5}(100 + V) \\\ \Rightarrow 100 \times 5 = 200 + 2V \\\ \Rightarrow 500 - 200 = 2V \\\ \therefore V = \dfrac{{300}}{2} = 150 \\\$
Therefore, the volume of bulb B is 150 mL.

Hence, the correct option is B.

Note – Whenever solving such types of questions then always write down the things given in the question, then use the formula as mentioned in the solution to find the value of the volume in mL, as the volume of A is 100 mL, and if the volume of B is V, then after opening the volume will be the sum of (100 + V), and also there is decrease in pressure. So, in this way we have found out the volume of B.