Question

18. A box of IL capacity is divided into two equal compartments by a thin partition which are filled with 2g
H2and 16g CH4, respectively. The pressure in each compartment is recorded as Patm. The total pressur
when partition is removed will be :
(A) P
(B) 2P
(C) P/2
(D) P/4​

Answer 1

Answer :  The correct options is, (A) P atm

Solution :

First we have to calculate the moles of  $H_2$ and $CH_4$.

$\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{2g}{2g/mole}=1mole$

$\text{Moles of }CH_4=\frac{\text{Mass of }CH_4}{\text{Molar mass of }CH_4}=\frac{16g}{16g/mole}=1mole$

Now we have to calculate the mole fraction of $H_2$ and $CH_4$.

Formula used :

$X_{H_2}=\frac{n_{H_2}}{n_{H_2}+n_{CH_4}}$

$X_{H_2}=\frac{1}{1+1}=\frac{1}{2}$

$X_{CH_4}=\frac{n_{CH_4}}{n_{H_2}+n_{CH_4}}$

$X_{CH_4}=\frac{1}{1+1}=\frac{1}{2}$

Now we have to calculate the partial pressure of $H_2$  and $CH_4$.

$p_{H_2}=X_{H_2}\times P_T$

$p_{H_2}=\frac{1}{2}\times P$

And,

$p_{CH_4}=X_{CH_4}\times P_T$

$p_{CH_4}=\frac{1}{2}\times P$

Now we have to calculate the total pressure when partition is removed.

$P_T=p_{H_2}+p_{CH_4}=\frac{1}{2}\times P+\frac{1}{2}\times P=P$

Therefore, the total pressure  when partition removed will be, P atm

Answer 2

Number of moles, n  H  2  =6/2=3  and n  CH  4  =16/16=1

P  H  2  = 3/ 3+1 ×P=  3 P/4

P  CH  4  = 1/1+3   ×P=P/4  

 p  total  =P  H  2  +P  CH  4   = P  

Hence, option A is correct