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An open vessel containing air is heated from 300 K TO 400 k . THE FRACTION OF AIR ORIGINALLY PRESENT WHICH GOES OUT OF IT IS :

  • Option 1)

    \frac{3}{4}

  • Option 2)

    \frac{1}{4}

  • Option 3)

    \frac{2}{3}

  • Option 4)

    \frac{1}{8}

 

Answers (1)

best_answer

As learnt in

Ideal Gas Law -

PV=nRT

- wherein

P - Pressure

V - Volume

n - No. of Moles

R - Gas Constant

T - Temperature

 

 PV = nRT , T1 = 300k , T2 = 400k

Since, it is an open vessel, pressure is constant and there is no change in the volume of vessel.

\therefore   P1 = P2  and V1 = V2

\frac{n_{1}}{n_{2}} = \frac{T_{2}}{T_{1}} = \frac{400}{300} = \frac{4}{3}

n_{2} = \frac{3}{4}n_{1}

Therefore, fraction of air that goes out = 1 - \frac{3}{4} = \frac{1}{4}


Option 1)

\frac{3}{4}

This option is incorrect

Option 2)

\frac{1}{4}

This option is correct

Option 3)

\frac{2}{3}

This option is incorrect

Option 4)

\frac{1}{8}

This option is incorrect

Posted by

Aadil

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