The KE of N molecule of O2 is x Joules at -123°C. Another sample of O2 at 27°C has a KE of 2x Joules. The latter sample contains

Question

The KE of N molecule of O2 is x Joules at -123°C. Another sample of O2 at 27°C has a KE of 2x Joules. The latter sample contains (A) N molecules of O2 (B) 2N molecules of O2 (C) N/2 molecules of O2 (D) N/4 molecule of O2

Tardigrade Admin · Accepted Answer

KE=(3/2)RT; T=-12+273=+150 L (3/2) × R × 150=3 × 8.314 ×75 =x J =225 × 8.314=x At 27°C=27+223=300 K KE for =2x J =(3/2)× 8.314× 300 N molecules ∴ x Joule =3 × 8.314 × 75 In both the cases x J correspond to N molecules

Q. The $K E$ of $N$ molecule of $O_{2}$ is x Joules at $- 12 3^{\circ} C$ . Another sample of $O_{2}$ at $2 7^{\circ} C$ has a $K E$ of $2 x$ Joules. The latter sample contains

$N$ molecules of $O_{2}$

$2 N$ molecules of $O_{2}$

$N /2$ molecules of $O_{2}$

$N /4$ molecule of $O_{2}$

Q. The KE of N molecule of O2​ is x Joules at −123∘C. Another sample of O2​ at 27∘C has a KE of 2x Joules. The latter sample contains

N molecules of O2​

2N molecules of O2​

N/2 molecules of O2​

N/4 molecule of O2​

KE=23​RT;T=−12+273=+150L 23​×R×150=3×8.314×75=xJ =225×8.314=x At 27∘C=27+223=300K KE for =2xJ=23​×8.314×300 N molecules ∴ x Joule =3×8.314×75 In both the cases x J correspond to N molecules

Q. The $K E$ of $N$ molecule of $O_{2}$ is x Joules at $- 12 3^{\circ} C$ . Another sample of $O_{2}$ at $2 7^{\circ} C$ has a $K E$ of $2 x$ Joules. The latter sample contains

$N$ molecules of $O_{2}$

$2 N$ molecules of $O_{2}$

$N /2$ molecules of $O_{2}$

$N /4$ molecule of $O_{2}$