Question

The temperature at which the average velocity of oxygen becomes equal to that of methane at 327ºC is (A) 627 ºC (B) 927 ºC (C) 1200 ºC (D) 127 ºC​

Answer 1

Answer:

The required temperature is 654°C.

Step-by-step-explanation:

We have given that,

The average velocity of oxygen is equal to that of methane at 327°C.

We know that,

$\displaystyle{\boxed{\pink{\sf\:V_{Avg}\:=\:\sqrt{\dfrac{8\:R\:T}{\pi\:M}}\:}}}$

Where,

$\displaystyle{\sf\:V_{Avg}\:=\:Average\:velocity}$

R = Universal gas constant = 8.314 J mol¯¹ K¯¹

T = Temperature

M = Molecular mass of gas

Now, from the given condition,

$\displaystyle{\boxed{\blue{\sf\:V_{Avg}\:(\:Oxygen\:)\:=\:V_{Avg}\:(\:Methane\:)\:}}}$

$\displaystyle{\implies\sf\:\sqrt{\dfrac{8\:R\:T_1}{\pi\:M_1}}\:=\:\sqrt{\dfrac{8\:R\:T_2}{\pi\:M_2}}}$

T₁, M₁ are temperature and molecular mass of oxygen.

T₂, M₂ are temperature and molecular mass of methane.

Molecular mass of oxygen O₂ = 2 * 16 = 32 g/mol

We know that,

Molecular formula of methane = CH₄

⇒ Molecular mass of methane = Molar mass of C + 4 * Molar mass of H

⇒ Molecular mass of methane = 12 + 4 * 1

⇒ Molecular mass of methane = 12 + 4

⇒ Molecular mass of methane = 16 g/mol

$\displaystyle{\implies\sf\:\sqrt{\dfrac{\cancel{8\:R\:}\:T_1}{\cancel{\pi\:}M_1}}\:=\:\sqrt{\dfrac{\cancel{8\:R\:}T_2}{\cancel{\pi\:}M_2}}}$

$\displaystyle{\implies\sf\:\sqrt{\dfrac{T_1}{32}}\:=\:\sqrt{\dfrac{T_2}{16}}}$

By squaring both sides, we get,

$\displaystyle{\implies\sf\:\dfrac{T_1}{32}\:=\:\dfrac{T_2}{16}}$

$\displaystyle{\implies\sf\:T_1\:=\:\dfrac{\cancel{32}\:\times\:T_2}{\cancel{16}}}$

$\displaystyle{\implies\sf\:T_1\:=\:2\:T_2}$

$\displaystyle{\implies\sf\:T_1\:=\:2\:\times\:327}$

$\displaystyle{\implies\:\underline{\boxed{\red{\sf\:T_1\:=\:654^{\circ}\:C\:}}}}$

∴ The required temperature is 654°C.