Search any question & find its solution
Question:
Answered & Verified by Expert
If the rms velocity of hydrogen gas at a certain temperature is $c,$ then the ras velocity of oxygen gas at the same temperature is
Options:
Solution:
1561 Upvotes
Verified Answer
The correct answer is:
$\frac{c}{4}$
$v_{\ln m}=\sqrt{\frac{3 R T}{M}}$
$R=$ gas constant
$T=$ absolute temperature
$M=$ mass of gas
$v_{\text {rms }}$ of $\mathrm{H}_{2}=\sqrt{\frac{3 R T}{M_{\mathrm{H}_{2}}}}$
$v_{\text {rms }}$ of $\mathrm{O}_{2}=\sqrt{\frac{3 R T}{M_{\mathrm{O}_{2}}}}$
$\frac{v_{\mathrm{rms}} \cdot \mathrm{H}_{2}}{v_{\mathrm{rma}} \cdot \mathrm{O}_{2}}=\sqrt{\frac{M_{O_{2}}}{M_{\mathrm{H}_{1}}}}$
Given $\quad v_{\operatorname{rms}} H_{2}=c$
$\Rightarrow \quad \frac{c}{v_{\operatorname{rms}} \mathrm{O}_{2}}=\sqrt{\frac{M_{0}}{M_{H_{1}}}}=\sqrt{\frac{32}{2}}=\sqrt{16}=4$
$\therefore \quad v_{rms}=\frac{C}{4}$
$R=$ gas constant
$T=$ absolute temperature
$M=$ mass of gas
$v_{\text {rms }}$ of $\mathrm{H}_{2}=\sqrt{\frac{3 R T}{M_{\mathrm{H}_{2}}}}$
$v_{\text {rms }}$ of $\mathrm{O}_{2}=\sqrt{\frac{3 R T}{M_{\mathrm{O}_{2}}}}$
$\frac{v_{\mathrm{rms}} \cdot \mathrm{H}_{2}}{v_{\mathrm{rma}} \cdot \mathrm{O}_{2}}=\sqrt{\frac{M_{O_{2}}}{M_{\mathrm{H}_{1}}}}$
Given $\quad v_{\operatorname{rms}} H_{2}=c$
$\Rightarrow \quad \frac{c}{v_{\operatorname{rms}} \mathrm{O}_{2}}=\sqrt{\frac{M_{0}}{M_{H_{1}}}}=\sqrt{\frac{32}{2}}=\sqrt{16}=4$
$\therefore \quad v_{rms}=\frac{C}{4}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.