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  • Give answer! Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27º C and contain 0.70 moles of H2 at 0.5 atm. One of the flask is then immersed into a bath kept at 127º C, while the other remains at 27&

Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at 27º C and contain 0.70 moles of H2 at 0.5 atm. One of the flask is then immersed into a bath kept at 127º C, while the other remains at 27º C.

The number of moles of H2 in flask 1 and flask 2 are:

  • Option 1)

     Moles in flask 1 = 0.4, Moles in flask 2 = 0.3  

  • Option 2)

     Moles in flask 1 = 0.2, Moles in flask 2 = 0.3

  • Option 3)

     Moles in flask 1 = 0.3, Moles in flask 2 = 0.2 

  • Option 4)

     Moles in flask 1 = 0.4, Moles in flask 2 = 0.2

 

Answers (1)

best_answer

As we learnt in 

Boyle’s Law -

At constant Temp and moles, pressure is inversely proportional to volume of gas.

- wherein

P_{1}V_{1}= P_{2}V_{2}     

 

Let vol. of each flask is 'V' L, Initially  

0.5 \times 2V = 0.7 \times 0.0821 \times 300

V = 17.24 L

Let T is final temperature of flask when pressure in each flask becomes equal, it happens. when

n_{total}=n_{1}+n_{2}

\frac{P\times \left ( 2V \right )}{R\times T}= \frac{PV}{R\times 400}= \frac{PV}{R\times 300}\Rightarrow \frac{2}{T}= \frac{1}{400}+\frac{1}{300}= \frac{3+4}{1200}= \frac{7}{1200}

T= \frac{2400}{7}= 342.85\: K

P \times 2 \times × 17.24 = n_{1} \times × 0.0821 \times × 400 \Rightarrow n_{1}= 0.3

0.5714 ×\times 17.24 = n_{2} ×\times 0.0821 ×\times 300\Rightarrow n_{2}= 0.4


Option 1)

 Moles in flask 1 = 0.4, Moles in flask 2 = 0.3  

Correct

Option 2)

 Moles in flask 1 = 0.2, Moles in flask 2 = 0.3

Incorrect

Option 3)

 Moles in flask 1 = 0.3, Moles in flask 2 = 0.2 

Incorrect

Option 4)

 Moles in flask 1 = 0.4, Moles in flask 2 = 0.2

Incorrect

Posted by

divya.saini

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