Number of molecules in a volume of 4 cm 3 of a perfect monoatomic gas at some temperature T and at a pressure of 2 cm of mercury is close to ? (Given, mean kinetic energy of a molecule (at T ) is 4 × 10-14 erg, g =980 cm / s 2, density of mercury .=13.6 g / cm 3)

Question

Number of molecules in a volume of 4 cm 3 of a perfect monoatomic gas at some temperature T and at a pressure of 2 cm of mercury is close to ? (Given, mean kinetic energy of a molecule (at T ) is 4 × 10-14 erg, g =980 cm / s 2, density of mercury .=13.6 g / cm 3) (A) 5.8 × 1018 (B) 5.8 × 1016 (C) 4.0 × 1018 (D) 4.0 × 1016

Tardigrade Admin · Accepted Answer

n =( PV / RT ), (3/2) kT =4 × 10-14 N =( PV / RT ) × Na =(2 × 13.6 × 980 × 4/(8)3 × 10-14)=3.99 × 1018

$5.8 \times 1 0^{18}$

$5.8 \times 1 0^{16}$

$4.0 \times 1 0^{18}$

$4.0 \times 1 0^{16}$

$n = \frac{P V}{RT}, \frac{3}{2} k T = 4 \times 1 0^{- 14}$
$N = \frac{P V}{RT} \times N a$
$= \frac{2 \times 13.6 \times 980 \times 4}{\frac{8}{3} \times 1 0 ^{- 14}} = 3.99 \times 1 0^{18}$

Q. Number of molecules in a volume of 4cm3 of a perfect monoatomic gas at some temperature T and at a pressure of 2cm of mercury is close to ? (Given, mean kinetic energy of a molecule (at T ) is 4×10−14 erg, g=980cm/s2, density of mercury =13.6g/cm3)

5.8×1018

5.8×1016

4.0×1018

4.0×1016

n=RTPV​,23​kT=4×10−14 N=RTPV​×Na =38​×10−142×13.6×980×4​=3.99×1018

$5.8 \times 1 0^{18}$

$5.8 \times 1 0^{16}$

$4.0 \times 1 0^{18}$

$4.0 \times 1 0^{16}$

$n = \frac{P V}{RT}, \frac{3}{2} k T = 4 \times 1 0^{- 14}$
$N = \frac{P V}{RT} \times N a$
$= \frac{2 \times 13.6 \times 980 \times 4}{\frac{8}{3} \times 1 0 ^{- 14}} = 3.99 \times 1 0^{18}$