Question

Ratio of $C _ { p }$ and $C _ { v }$ of a gas 'X' is 1.4. The number of atoms of the gas 'X' present in 11.2 litres of it at N.T.P. is

• A. $6.02 \times 10 ^ { 23 }$
• B. $1.2 \times 10 ^ { 24 }$
• C. $3.01 \times 10 ^ { 23 }$
• D. $2.01 \times 10 ^ { 23 }$

Answer 1

Answer: (A)

$6.02 \times 10 ^ { 23 }$

Answer 2

Since $\frac { C _ { P } } { C _ { V } } = 1.4$, the gas should be diatomic.

If volume is 11.2 lt then, no. of moles = $\frac { 1 } { 2 }$

no. of molecules = $\frac { 1 } { 2 }$× Avagadro’s No.

no. of atoms = 2 × no. of molecules

2 × $\frac { 1 } { 2 }$× Avagadro’s No.

$= 6.0223 \times 10 ^ { 23 }$