Question

Total number of molecules in 'A' is ($N_A$ = Avagadro number)

• A. $\frac{125}{64}{N}_A$
• B. $3.125N_A$
• C. $\frac{125}{28}{N}_A$
• D. $31.25N_A$

Answer 1

Answer: (B)

$3.125N_A$

Since the temperature rise and heat supplied to both the container is same.

$Q=n_A{C}_{VA}\Delta T=n_B{C}_{VB}\Delta T$ $\Rightarrow n_A{C}_{VA}=n_B{C}_{VB}-\left(1\right)$

also, $\Delta P_{A}V=n_{A}R\Delta T-\left(2\right)$

and, $\Delta P_{B}V=n_{B}R\Delta T-\left(3\right)$

Dividing $\left(2\right)$ by $\left(3\right)$,

$n_A/n_B=\Delta P_A/\Delta P_B=2.5/1.5=5/3$

$\Rightarrow n_A=5n_B/3$

From equation $\left(1\right)$

$C_{VB}/C_{VA}=5/3=\left(5R/2\right)/\left(3R/2\right)$

$C_{VB}=5R/2$, so $B$ is diatomic.

$C_{VA}=3R/2$, so $A$ is Mono-atomic.

Since $n_A/n_B=5/3-\left(1\right)$

mass of the container is $100g$

Mass of $A$ is $225-100=125$

Let the molecular mass be $M_{A}and{M}_B$

so, $n_A=125/M_A,n_B=60/M_B-\left(2\right)$

From $\left(1\right)and\left(2\right)$

$5M_B=4M_A$

Mass of argon is $40$, ozygen is $32$ which satisfy the above relation.

So, $A$ contains Argon and $B$ contains oxygen.

No. of molecules in 'A' = $n{N}_A$
$=\frac{125}{40}{N}_A$
$=3.125N_A$ (Since n= $\frac{125}{40}$ for A)