The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300 K is: (Atomic masses: He = 4 u, 0 = 16 u) The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300 K is: (Atomic masses: He = 4 u, 0 = 16 u)

Question

The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300 K is: (Atomic masses: He = 4 u, 0 = 16 u) The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300 K is: (Atomic masses: He = 4 u, 0 = 16 u) (A) 300 K (B) 600 K (C) 1200 K (D) 2400 K

Tardigrade Admin · Accepted Answer

(Urms)o2 =(Urms)He ⇒ (3RTo2/32)=(3R THe/4) ⇒ To2 =8×300=2400 K

$300 K$

$600 K$

$1200 K$

$2400$ K

$(U_{r m s})_{o_{2}} = (U_{r m s})_{He}$

$\Rightarrow \frac{3 RT o _{2}}{32} = \frac{3 R T _{He}}{4}$

$\Rightarrow T_{o_{2}} = 8 \times 300 = 2400 K$

Q. The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300K is : (Atomic masses : He=4u,0=16u) The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300K is : (Atomic masses : He=4u,0=16u)

300K

600K

1200K

2400K

(Urms​)o2​​=(Urms​)He​ ⇒323RTo2​​=43RTHe​​ ⇒To2​​=8×300=2400K

$300 K$

$600 K$

$1200 K$

$2400$ K

$(U_{r m s})_{o_{2}} = (U_{r m s})_{He}$

$\Rightarrow \frac{3 RT o _{2}}{32} = \frac{3 R T _{He}}{4}$

$\Rightarrow T_{o_{2}} = 8 \times 300 = 2400 K$