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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).

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Valve- V will be the second valve which open.

Select the correct options(s).

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

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

answer is B, C.

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

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