Ratio of Cp and Cv of a gas 'X' is 1.4. The number of atoms of the gas 'X' present in 11.2 litres of it at NTP is

Question

Ratio of Cp and Cv of a gas 'X' is 1.4. The number of atoms of the gas 'X' present in 11.2 litres of it at NTP is (A) 6.02 × 1023 (B) 1.2 × 1024 (C) 3.0× 1023 (D) 2.01 × 1023

Tardigrade Admin · Accepted Answer

Since (CP/CV)=1.4, the gas should be diatomic If volume is 11.2 L, then no. of moles =(1/2) ∴ No. of molecules =(1/2) × Avogadroâs No No. of atoms =2 × no. of molecules 2 × (1/2) × Avogadroâs No. =6.0223× 1023

Q. 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 $NTP$ is

$6.02 \times 1 0^{23}$

$1.2 \times 1 0^{24}$

$3.0 \times 1 0^{23}$

$2.01 \times 1 0^{23}$

Since $\frac{C _{P}}{C _{V}} = 1.4$ , the gas should be diatomic
If volume is 11.2 L, then no. of moles $= \frac{1}{2}$
$∴$ No. of molecules $= \frac{1}{2}$ $\times$ Avogadro’s No
No. of atoms $= 2 \times$ no. of molecules
$2 \times \frac{1}{2} \times$ Avogadro’s No. $= 6.0223 \times 1 0^{23}$

Q. Ratio of Cp​ and Cv​ of a gas ′X′ is 1.4. The number of atoms of the gas ′X′ present in 11.2 litres of it at NTP is

6.02×1023

1.2×1024

3.0×1023

2.01×1023

Since CV​CP​​=1.4, the gas should be diatomic If volume is 11.2 L, then no. of moles =21​ ∴ No. of molecules =21​ × Avogadro’s No No. of atoms =2× no. of molecules 2×21​× Avogadro’s No. =6.0223×1023

Q. 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 $NTP$ is

$6.02 \times 1 0^{23}$

$1.2 \times 1 0^{24}$

$3.0 \times 1 0^{23}$

$2.01 \times 1 0^{23}$