At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal to that of oxygen molecules at 27 degree Celsius?

Answer (Detailed Solution Below)

Option 3 : 6495°C

Concept:

Root Mean Square Speed

• It is defined as the square root of the mean of squares of the speed of different molecules.
• The root-mean-square speed takes into account both molecular weight and temperature, two factors that directly affect the kinetic energy of a material.
• The rms speed of any homogeneous gas sample is given by:

Where R = universal gas constant, T = temperature and M = Molecular mass

Calculation:

Let velocity of oxygen = Vo

Let velocity of hydrogen = Vh

Molecular weight of oxygen Mo = 32

Molecular weight of hydrogen Mh = 2

Let the required temperature be T0 = 

Temperature of hydrogen TH =  150 °C = 150 + 273 = 423 K

Now

Equating

Squaring and cancling constant terms 

we get

⇒ To = 16 Th

⇒ To = 16 × 423 K = 6768 K

This temperature in °C will be 6768  -  273 = 6495°C

So, the correct option is 6495°C.

Mistake Points

Aspirants make take the temperature in °C. It should be taken in Kelvin, the SI unit for any calculation purpose. 

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Text Solution

`20 K``80 K``-73 K``3 K`

Answer : A

Solution : For oxygen `V_(O_2) = sqrt((3RT_(O_2))/(M_(O_2)))` and <br> For hydrogen, `v_(H_2) = sqrt(3R(T_(H_2))/(M_(H_2)))` <br> According to problem `= sqrt((3RT_(O_2))/(M_(O_2))) = sqrt(3R(T_(H_2))/(M_(H_2)))` <br> `implies (T_(O_2))/(M_(O_2)) = (T_(H_2))/(M_(H_2)) implies (47+273)/(32) = (T_(H_2))/(M_(H_2))` <br> `implies T_(H_2) = (320)/(32)xx2 = 20K`.