Question 1: Thermodynamics & KTG

Question

A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially the bottle is sealed at atmosphere pressure p
0
= 10
5
Pa so that the volume of the trapped air is v
0
= 3.3 cc. When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure p
0
+ $\Delta$p without changing its orientation. At this pressure, the volume of the trapped air is v
0
$-$ $\Delta$v. Let $\Delta$v = X cc and $\Delta$p = Y $	imes$ 10
3
Pa.
The value of Y is _______________.
![](https://imagex.cdn.examgoal.net/1kuffm4qb/db4ec4a9-cf8d-4ee9-b52f-98ee4071e22b/23d18c30-2699-11ec-abd8-3b6ac251d2e9/file-1kuffm4qc.png)

Accepted Answer

When the tube just sinks/floats, then average density = density of water
${{Mass} \over {total\,volume}}$ = 1 gm/cc
$ \Rightarrow {{5gm} \over {total\,volume}}$ = 1 gm/cc
$\Rightarrow$ Total volume = 5 cc
$\Rightarrow$ Volume of tube + final volume of air in the tube = 5 cc
$ \Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$
$\Rightarrow$ V
f
= 5 $-$ 2 = 3 cc
$\Rightarrow$ $\Delta$V = 0.3 cc
For isothermal process,
${p_i}{V_i} = {p_f}{V_f} \Rightarrow {p_f} = {10^5} 	imes {{3.3} \over 3}$
p
f
= 1.1 $	imes$ 10
5
p
f
$-$ p
i
= 1.1 $	imes$ 10
5
$-$ 10
5
= 0.1 $	imes$ 10
5
= 10 $	imes$ 10
3
Pa $\Rightarrow$ Y = 10

Thermodynamics & KTG

Integer / Numeric Answer Type