## haridas_mandal 3 years ago two glass bulbs A &B of volume V , 2V respectively are connected by a narrow capilary tube. The bulbs contain gas at temp T and Pr P. Now the temp of bulb A is doubled and that of B is tripled. The final number of moles of gas in bulb A is : 9 options 3/2 PV/RT, 5/7 PV/RT, 12/7 PV/RT & 9/7 PV /RT.

1. haridas_mandal

Pl explain

2. haridas_mandal

It is 9/7 PV/RT as per the printed question paper in front of me..

3. experimentX

hold on ... I may be wrong.

4. experimentX

the volume is same and since the tubes are connected the pressure is also same. the total no of moles is ${3 PV \over RT}$

5. experimentX

now the gases are heated with temperature $$2T$$ and $$3T$$, the new pressure is be $$P_1$$ Now, ${P_1 V \over R 2T} + {P_1 2V \over R 3T} = {3PV \over RT}\\ P_1 = {18P \over 7}$ The number of moles in tube A is given by $${P_1 V \over R 2T}$$ ${P_1 V \over R 2T} = {18 P V \over 7 *2 RT} = {9 \over 7 }..$

6. experimentX

I guess the last option is correct.

7. haridas_mandal

I think that's the correct answer. Thanks