Question

If the ratio of $C_P$ and $C_V$ for a gas is 1.4. The number of atoms present in 11.2 litre of it at STP is:

• A. $6.02\times {10}^{23}$
• B. $1.2\times {10}^{23}$
• C. $3.01\times {10}^{23}$
• D. $12.04\times {10}^{23}$

Answer 1

The correct option is A $6.02\times {10}^{23}$

The ratio of the $C_p$and $C_v$ of a gas gives the atomicity of the gases if,$\frac{C_p}{C_v}=1.66$ (monoatomic)

$\frac{C_p}{C_v}=1.40$ (diatomic) And

$\frac{C_p}{C_v}=1.03$ (Tri atomic)

Since, Here $\frac{C_p}{C_v}=1.4$ so the gases diatomic like

$H_2,N_2$ etc

Now, We know that,at NTP, 1mole of gas is $=22.4l$

So, 1 mole will have $6.02\times {10}^{23}$ molecules of gas $6.02\times {10}^{23}\times 2$ atoms of gas.

So $0.5$ mole of gas will be

$=6.02\times {10}^{23}\times 2\times 0.5$

$=6.02\times {10}^{23}$