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QJEEM21Ch823c: JEE Main 2021 Chemistry - Class 11 JEEM21Ch823c Solution: Two flasks I and II shown below are connected by a valve of…

Two flasks I and II shown below are connected by a valve of negligible volume.

JEE Main 2021 (Online) 27th August Evening Shift Chemistry - Thermodynamics Question 90 English
When the valve is opened, the final pressure of the system in bar is x ×\times 10 −- 2. The value of x is __________. (Integer answer)

[Assume - Ideal gas; 1 bar = 105 Pa; Molar mass of N2 = 28.0 g mol −- 1; R = 8.31 J mol −- 1 K −- 1]

Step-by-Step Verified Solution

Applying; (nI + nII)initial = (nI + nII)final

⇒\Rightarrow Assuming the system attains a final temperature of T (such that 300 < T < 60)

⇒\Rightarrow (HeatlostbyN2ofcontainerI)=(HeatgainedbyN2ofcontainerII)

⇒\Rightarrow n1Cm(300 −- T) = nIICm(T −- 60)

⇒(2.828)(300−T)=0.228(T−60)\Rightarrow \left( {{{2.8} \over {28}}} \right)(300 - T) = {{0.2} \over {28}}(T - 60)

⇒\Rightarrow 14(300 −- T) = T −- 60

⇒\Rightarrow (14×300+60)15=T{{(14 \times 300 + 60)} \over {15}} = T

⇒\Rightarrow T = 284 K (final temperature)

⇒\Rightarrow If the final pressure = P

⇒\Rightarrow (nI + nII)final = (3.028)\left( {{{3.0} \over {28}}} \right)

⇒\Rightarrow PRT(VI+VII)=3.0 gm28 gm/mol{P \over {RT}}({V_I} + {V_{II}}) = {{3.0\,gm} \over {28\,gm/mol}}

P=(328mol)×8.31Jmol−K×284K3×10−3m3×10−5barPaP = \left( {{3 \over {28}}mol} \right) \times 8.31{J \over {mol - K}} \times {{284K} \over {3 \times {{10}^{ - 3}}{m^3}}} \times {10^{ - 5}}{{bar} \over {Pa}}

⇒\Rightarrow 0.84287 bar

⇒\Rightarrow 84.28 ×\times 10 −- 2 bar

⇒\Rightarrow 84

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