Question

3) At 20°C ,two balloons of equal volume and porosity are filled to pressure of 2atm ,one with 14Kg N₂ and other with 1Kg of H₂.the N₂ balloon leaks to a pressure of ½ atm in 1 hr.if it take t min for H₂ balloon to reach a pressure of ½ atm then calculate t/4

Answer 1

Solution: According to the initial situation, we get W2. As P1/P2=W1/W2, => 2/1/2= 14/W2. Therefore,

W2 =14/4 kgN^2. Weight of N2 diffused in 1 hour=14-14/4 = 21/2kg. Similarly, for H2, W2=1/4kgH^2.

Weight of H2 diffused=1-1/4=3/4. For diffusion of N2 and H2, tH2=1hr. We know, (WH2/tH2)*(tN2/WN2)=(MH2/MN2)^1/2.

Hence on calculating, we find t= 60 minutes.

Answer 2

Explanation:

Dear Student,

For N2 : Initially, P1=2 atm w1= 14 kg ⎜diffusion time 1 hr| After diffusion, P2=12atm. w2=?Since at constant V and temperature, for a gas. P α w

Therefore, P1/P2=w1/w2

or, 2/0.5=14/w2or,

w2=14/4kg

Therefore, weight of N2 diffused in 1 hr=14−14/4=21/2 kg

Similarly for H2.

Initially P1=2 atm w1=1 kg ⎜diffusion time t hr|

After diffusion, P2=1/2atm w2=?Again, P1/P2=w1/w2or,

2/0.5=1/w2or,

w2=14kg

Therefore, weight of H2 diffused=1−1/4=3/4

after this use the diffusion formula. the answer comes out to be 16 min.

so t/4= 4