Question

Two closed vessels of the equal volume contain air at $105KPa$ at $300K$ and are connected through a narrow tube. If one of the vessels is now maintained at $300K$ and the other at $400K$ then the pressure becomes.
(A) $120KPa$
(B) $105KPa$
(C) $150KPa$
(D) $300KPa$

Answer 1

Hint : To solve this question we have to know about pressure. We know that, Pressure (symbol: p or P) is the force applied perpendicular to the surface of an object per unit area over which that force is distributed. Gauge pressure (also spelled gage pressure) is the pressure relative to the ambient pressure. Various units are used to express pressure. Some of these derive from a unit of force divided by a unit of area; the SI unit of pressure, the pascal (Pa)

Complete Step By Step Answer:
In this case, the initial pressure is equal to P and the final pressure is equal to P’.
Since $(P/T)$ is constant.
Considering the system of two spheres of equal volume, we get,
$\dfrac{P}{{300}} + \dfrac{P}{{300}} + \dfrac{{P'}}{{300}} + \dfrac{{P'}}{{400}}$
Here, $P = 105KPa$
Solving these above equations, we will get, P’ $= 120KPa$
So, the right answer will be A.

Note :
We also have to know that it is easier to hammer a sharp pain than to hammer a blunt pin. This is because the area at the end of the sharp pin is smaller than the area at the end of a blunt pin. This leads to an increase in pressure leading to hammering the sharp pin easily. We have to keep these also in our mind to solve this question correctly.