Question 175: Thermodynamics & KTG

Question

The r.m.s. speed of oxygen molecules at 47 °C is equal to that of the hydrogen molecules kept at _________ °C. (Mass of oxygen molecule/mass of hydrogen molecule = 32/2)

Accepted Answer

According to the Kinetic Theory of Gases, the r.m.s speed of an ideal gas is given by :
$$ \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} $$
Where T is the absolute temperature and M is the molecular mass (in kg ).
It is given that the r.m.s. speed of oxygen $\left(\mathrm{O}_2ight)$ at $\mathrm{T}_{\mathrm{O}_2}$ is equal to the r.m.s. speed of hydrogen $\left(\mathrm{H}_2ight)$ at $\mathrm{T}_{\mathrm{H}_2}$.
$$ \mathrm{V}_{\mathrm{rms}\left(\mathrm{O}_2ight)}=\mathrm{V}_{\mathrm{rms}\left(\mathrm{H}_2ight)} $$
$$\Rightarrow $$ $\sqrt{\frac{3 \mathrm{RT}_{\mathrm{O}_2}}{\mathrm{M}_{\mathrm{O}_2}}}=\sqrt{\frac{3 \mathrm{RT}_{\mathrm{H}_2}}{\mathrm{M}_{\mathrm{H}_2}}}$
$$\Rightarrow $$ $$ \frac{\mathrm{T}_{\mathrm{O}_2}}{\mathrm{M}_{\mathrm{O}_2}}=\frac{\mathrm{T}_{\mathrm{H}_2}}{\mathrm{M}_{\mathrm{H}_2}} $$
Temperature of oxygen is $\mathrm{T}_{\mathrm{O}_2}=47+273=320 \mathrm{~K}$.
Molar masses are $\mathrm{M}_{\mathrm{O}_2}=32 	imes 10^{-3} \mathrm{~kg}$ and $\mathrm{M}_{\mathrm{H}_2}=2 	imes 10^{-3} \mathrm{~kg}$.
$$ \frac{320}{32}=\frac{\mathrm{T}_{\mathrm{H}_2}}{2} $$
$$\Rightarrow $$ $10=\frac{\mathrm{T}_{\mathrm{H}_2}}{2} \Rightarrow \mathrm{~T}_{\mathrm{H}_2}=20 \mathrm{~K}$
$$\Rightarrow $$ $\mathrm{T}_{\mathrm{H}_2}=20-273$
$$\Rightarrow $$ $$ \mathrm{T}_{\mathrm{H}_2}=-253^{\circ} \mathrm{C} $$
Therefore, the rms speed of oxygen molecules at $47^{\circ} \mathrm{C}$ is equal to that of the hydrogen molecules kept at $-253^{\circ} \mathrm{C}$.
Hence, the correct option is (B).

Answer

-100

Answer

-253

Answer

-20

Answer

-235

Thermodynamics & KTG

The r.m.s. speed of oxygen molecules at 47 °C is equal to that of the hydrogen molecules kept at _________ °C. (Mass of oxygen molecule/mass of hydrogen molecule = 32/2)

Choose the correct answer: