A vessel contains two non-reactive gases; neon (monoatomic) and oxygen (diatomic). The ratio of their partial pressures is 3: 2 . Estimate the ratio of their number of molecules. Atomic mass of Ne =20.2 u and Atomic mass of O 2=32 u

Question

A vessel contains two non-reactive gases; neon (monoatomic) and oxygen (diatomic). The ratio of their partial pressures is 3: 2 . Estimate the ratio of their number of molecules. Atomic mass of Ne =20.2 u and Atomic mass of O 2=32 u (A) (3/2) (B) (2/3) (C) (20.2 × 2/32 × 3) (D) (20.2 × 3/32 × 2)

Tardigrade Admin · Accepted Answer

For a given volume of the vessel and constant temperature P propto μ. μ= no. of moles. (P1/P2)=(μ1/μ2)=((N1 / NA)/(N2 / NA)) ⇒ (N1/N2)=(P1/P2)=(3/2)

Q. A vessel contains two non-reactive gases; neon (monoatomic) and oxygen (diatomic). The ratio of their partial pressures is 3: 2 . Estimate the ratio of their number of molecules. Atomic mass of $N e = 20.2 u$ and Atomic mass of $O_{2} = 32 u$

$\frac{3}{2}$

$\frac{2}{3}$

$\frac{20.2 \times 2}{32 \times 3}$

$\frac{20.2 \times 3}{32 \times 2}$

For a given volume of the vessel and constant temperature $P \propto μ$ .
$μ =$ no. of moles.
$\frac{P _{1}}{P _{2}} = \frac{μ _{1}}{μ _{2}} = \frac{( N _{1} / N _{A} )}{( N _{2} / N _{A} )}$
$\Rightarrow \frac{N _{1}}{N _{2}} = \frac{P _{1}}{P _{2}} = \frac{3}{2}$

Q. A vessel contains two non-reactive gases; neon (monoatomic) and oxygen (diatomic). The ratio of their partial pressures is 3: 2 . Estimate the ratio of their number of molecules. Atomic mass of Ne=20.2u and Atomic mass of O2​=32u

23​

32​

32×320.2×2​

32×220.2×3​