Temperature at which most probable speed of $O_{2}$ becomes equal to root mean square speed of $N_{2}$ is:[Given: $N_{2}$ at 427 $^{\circ}$ C]

Open in App

Solution

Root mean square speed is $v_{r} = \sqrt{\frac{3 R T}{M}}$
For $N_{2}, M = 14 \times 2 = 28$
$v_{r} = \sqrt{\frac{3 R T}{28}} = \sqrt{\frac{3 \times 8.314 \times (427 + 273)}{28}} =$
Most probable speed is $v_{r} = \sqrt{\frac{2 R T}{M}}$
For $O_{2}, M = 16 \times 2 = 32$
$v_{M} = \sqrt{\frac{2 R T}{M}} = \sqrt{\frac{2 \times 8.314 \times T}{32}}$
Now,
$\sqrt{\frac{3 \times 8.314 \times (427 + 273)}{28}} = \sqrt{\frac{2 \times 8.314 \times T}{32}}$
$\Rightarrow T = 1200 K$

Suggest Corrections

Similar questions

O_{2}

N_{2}

N_{2}

427^{0} C

O_{2}

O_{2}

N_{2}

N_{2}

440 K

S O_{2}

O_{2}

300 K

Join BYJU'S Learning Program