NEET-XII-Physics
24: Kinetic Theory of Gases
- #17The mean speed of the molecules of a hydrogen sample equals the mean speed of the molecules of a helium sample. Calculate the ratio of the temperature of the hydrogen sample to the temperature of the helium sample.Ans : Mean velocity is given by
`` {V}_{avg}=\sqrt{\frac{8RT}{\pi M}}``
Let temperature for H and He respectively be T​1 and T2, respectively.
For hydrogen:
MH = 2g = 2`` \times ``10-3 kg
For helium:
MHe= 4 g = 4`` \times ``10-3 kg
Now,
A/q`` \sqrt{\frac{8R{T}_{1}}{\pi {M}_{H}}}=\sqrt{\frac{8R{T}_{2}}{\pi {M}_{He}}}``
`` \Rightarrow \sqrt{\frac{8R{T}_{1}}{2\times {10}^{-3}\pi }}=\sqrt{\frac{8R{T}_{2}}{\pi \times 4\times {10}^{-3}}}``
`` \Rightarrow \sqrt{\frac{{T}_{1}}{2}}=\sqrt{\frac{{T}_{2}}{4}}``
`` \Rightarrow \frac{{T}_{1}}{{T}_{2}}=\frac{1}{2}``
`` \Rightarrow {T}_{1}:{T}_{2}=1:2``