The temperature at which root mean square velocity of molecules of helium is equal to root mean square velocity of hydrogen at NTP is

Question

The temperature at which root mean square velocity of molecules of helium is equal to root mean square velocity of hydrogen at NTP is (A) 273?C (B) 273 K (C) 546?C (D) 844 K

Tardigrade Admin · Accepted Answer

vrms=√(3RT/M) rms velocity of He and H2 is same. i.e., (TH2/MH2)=(THe/MHe) or (273/2)=(THe/4) or (273/2)=(THe/4) =546-273 =273° C

Q. The temperature at which root mean square velocity of molecules of helium is equal to root mean square velocity of hydrogen at NTP is

273?C

273 K

546?C

844 K

$v_{r m s} = \frac{3 RT}{M}$ rms velocity of He and $H_{2}$ is same. i.e., $\frac{T _{H_{2}}}{M _{H_{2}}} = \frac{T _{He}}{M _{He}}$ or $\frac{273}{2} = \frac{T _{He}}{4}$ or $\frac{273}{2} = \frac{T _{He}}{4}$ $= 546 - 273$ $= 273^{\circ} C$

Q. The temperature at which root mean square velocity of molecules of helium is equal to root mean square velocity of hydrogen at NTP is

273?C

273 K

546?C

844 K

vrms​=M3RT​​ rms velocity of He and H2​ is same. i.e., MH2​​TH2​​​=MHe​THe​​ or 2273​=4THe​​ or 2273​=4THe​​ =546−273 =273∘C

$v_{r m s} = \frac{3 RT}{M}$ rms velocity of He and $H_{2}$ is same. i.e., $\frac{T _{H_{2}}}{M _{H_{2}}} = \frac{T _{He}}{M _{He}}$ or $\frac{273}{2} = \frac{T _{He}}{4}$ or $\frac{273}{2} = \frac{T _{He}}{4}$ $= 546 - 273$ $= 273^{\circ} C$