A container fitted with frictionless massless piston consists of five valves-I, II, ill, IV and V. These valves open automatically if pressure exceed over 1 .5, 2.2, 2.5, 4.4 and 4.8 atm respectively. Under the given initial conditions (mentioned in· given diagram) system is in state of equilibrium. Piston is now pressed in downward direction very slowly. ;
[Note : Consider the diameter of valve tube negligible and temperature remains constant.]

Select the correct options(s).

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

Select the correct options(s).

As the piston crosses the valve which will be opened first, the remaining number of moles in container are $\frac{5}{3}$ ,

Valve- V will be the second valve which open.

Number of moles will zero as piston crosses Valve-V.

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

At valve I position of piston

$1 \times 60 = P \times 45$

$\Rightarrow P = 1.33 < 1.5$

So, valve I does not open.

At valve II position of piston,

$1 \times 60 = P \times 30$

$\Rightarrow P = 2 atm < 2.2$

So, valve II does not open.

At valve III position of piston,

$1 \times 60 = P \times 20$

$\Rightarrow P = 3 atm > 2.5$

So, valve III opens till P becomes 2.5 atm.

At valve IV position of piston,

$2.5 \times 20 = P \times 15$

$\Rightarrow P = 3.33 < 4.4$

Hence, valve V opens second till pressure becomes 4.8 atm.

After opening of valve III, number of moles

remaining $= \frac{2.5 \times 20 \times 821 \times 10^{- 3}}{0.0821 \times 300} = \frac{5}{3}$ .

Watch 3-min video & get full concept clarity