Question

Initially bulb $"A"$ contained oxygen gas at ${27}^{\circ }C$ and $950mm$ of $Hg$ and bulb $"B"$ contained neon gas at ${27}^{\circ }C$ and $900mm$ of $Hg$. These bulbs are connected by a narrow tube of negligible volume equipped with a stopcock and gases were allowed to mix-up freely. Then obtain pressure in the combined system was found to be $910mm$ of $Hg$. If volume of bulb B was measured to be 10 L.

Then find the mass of oxygen gas present initially in bulb "A".

Answer 1

The ideal gas equation is $PV=nRT$.

Let the volume of bulb A be V ml.

For bulb A, $950V=nRT$......(1)

For bulb B, $900\times 10=n^{\prime }RT$......(2)

When the stopcock is opened, $910(10+V)=(n+n^{\prime })RT=nRT+n^{\prime }RT$......(3)

Substitute (1) and (2) into (3):

$9100+910V=950V+9000$

$100=40V$

$V=2.5L$

Substitute in (1):

The number of moles of oxygen in bulb A is:

$n=\frac{950\times 2.5}{760\times 0.0821\times 300}=0.1268$

The mass of oxygen in bulb A is $w=32\times 0.1268=4g$.