Question

Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at ${27}^{\circ }C$ and contain 0.70 moles of $H_2$ at 0.5 $atm$. One of the flask is then immersed into a bath kept at ${127}^{\circ }C$, while the other remains at ${27}^{o}C$. The number of moles of $H_2$ in flask 1 and flask 2 are :

• A. Moles in flask 1 = 0.4, Moles in flask 2 = 0.3
• B. Moles in flask 1 = 0.2. Moles in flask 2 = 0.3
• C. Moles in flask 1 = 0.3. Moles in flask 2 = 0.2
• D. Moles in flask 1 = 0.4, Moles in flask 2 = 0.2

Answer 1

The correct option is A Moles in flask 1 = 0.4, Moles in flask 2 = 0.3

$PV=nRT$

$0.5V=0.7R300$

so $V=420R$

In 2nd case, the pressure will be the same in both flasks and the sum of moles of gas will be 0.7. But volume will be half $420R/2=210R$

Flask 1-

$PV=aRT$

$P210R=aR300\Rightarrow a=0.7P$

Flask2-

$PV=bRT$

$P210R=bR400\Rightarrow b=0.525P$

$a+b=1.225P=0.7$

$P=0.571atm$

$a=0.7\times 0.571=0.399=0.4moles$

$b=0.525\times 0.571=0.299=0.3moles$

Hence, the correct option is $A$