1.

The system shown in the figure is in equilibrium, where A and B are isomeric liquids and form an ideal solution at TK.standard vapour pressure of A and B are P_A^@ and P_B^@ respectively at TK.We collect the vapour of A and B in two container of volume V, first container is maintained at 2TK and second container is maintained at (3T)/2K.At the temperature greater than TK, both A and B exist in only gaseous form.Total vapour pressure of the system at TK is given as P=P_A^@X_A+P_B^@X_B where X_A and X_B are the mole fraction of A and B in a liquid mixture. In container (1) We assume that collected gases behave ideally at 2TK and there can take place isomerisation reaction in which A converted into B by first order kinetics Reaction is given as AoversetAtoB In container (II) At the given temperature (3T)/2,A and B are ideal in nature and non mixing in nature.A small pin hole is made into container.We can determine the initial rate of effusion of both gases in vaccum by the expression r=(kp)/sqrtm Where P= pressure difference between system and surrounding K=positibe constant M=Molecular weight of the gas. Vapour is collected and passed into a container of volume 8.21 lit, maintainer at 100 K and after 5 min number of mole of B=8/3.Then calculate pressure develop into the container after two half lives.

Answer»

2 ATM
4 atm
1 atm
0.5 atm

Solution :`{:(,AtoB),(t=0,8/3" "4/3),(t=5"min",4/3" "8/3):}`
Now after two HALF life
`n_A=2/3` and `n_B=10/3`
`P_T V=(n_A+n_B)RT`
`P_Txx8.21=12/3xx0.082xx100 " "implies P_T=4 "atm"`


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