Question

The temperature at which the average speed of oxygen molecules is double that of the same molecules at 0c

Answer 1

To find:

The temperature at which the average speed of oxygen molecules is double that of the same molecules at 0°C.

Calculation:

Let the required temperature be T ;

$\therefore \: (v1) =2( v2)$

$= > \sqrt{ \dfrac{8RT}{\pi m} } = 2 \times \sqrt{ \dfrac{8R(0 + 273)}{\pi m} }$

$= > \sqrt{ \dfrac{ \cancel{8R}T}{ \cancel{\pi m}} } = 2 \times \sqrt{ \dfrac{ \cancel{8R}(0 + 273)}{ \cancel{\pi m}} }$

$= > \sqrt{T} = 2 \times \sqrt{ 273}$

Squaring both sides;

$= > T = 4 \times 273$

$= > T = 1092 \: K$

$= > T = {819}^{ \circ} C$

So, final answer is:

$\boxed{ \sf{ T = {819}^{ \circ} C}}$

Answer 2

Given:

The average speed of oxygen molecules is double that of the same molecules at 0c

To find:

The temperature at which the average speed of oxygen molecules is double that of the same molecules at 0c

Solution:

From given, we have,

T₁ = 0 °C = 273 K

The formula used to calculate the average speed of the gas is given by,

v{avg} = √[πRT/8m]

where, m = nN

nN represents the number of molecules.

From given, we have,

√[8RT₁/πm] = √[8RT₂/π(2m)]

squaring on both the sides, we get,

8RT₁/πm = 8RT₂/π(2m)

upon further solving we get,

T₁ = T₂/2

⇒ T₂ = 2T₁

⇒ T₂ = 2 × 273

∴ T₂ = 546 K

The temperature at which the average speed of oxygen molecules is double that of the same molecules at 0c is 546 K.