KTGHard
Question
The r.m.s speed of oxygen molecules at $47^{\circ}C$ is equal to that of the hydrogen molecules kept at $\_\_\_\_$ $\ ^{\circ}C$. (Mass of oxygen molecule/mass of hydrogen molecule=32/2)
Options
A.-235
B.-100
C.-253
D.-20
Solution
$V_{rms} = \sqrt{\frac{3R\text{ }T}{M}}$
$${V_{{rmsO}_{2}} = V_{{rmsH}_{2}} }{T_{O_{2}} = 273 + 47 = 320\text{ }K }{\sqrt{\frac{3{RT}_{O_{2}}}{M_{O_{2}}}} = \sqrt{\frac{3{RT}_{H_{2}}}{M_{H_{2}}}} }{\frac{T_{2}}{M_{O_{2}}} = \frac{T_{H_{2}}}{M_{H_{2}}} }{\frac{320}{32} = \frac{T_{H_{2}}}{2} }{T_{H_{2}} = 20\text{ }K }{T_{H_{2}} = - 253^{\circ}C}$$
Create a free account to view solution
View Solution FreeMore KTG Questions
Refer to fig. Let ᐃU1 and ᐃU2 be the changes in internal energy of the system in the processes A and B then...The value of CV for one mole of Neon gas is-...An ideal gas can be expanded from an initial state to a certain volume through two different processes (i) PV2 = constan...Consider two boxes containing ideal gases A and B such that their temperatures, pressures and number densities are same....Temperature at which the velocity of sound in O2 is the same as that on N2 at 27oC is approximately...