A) \[\frac{1}{2}\sqrt{15}\,km{{s}^{-1}}\]
B) \[\frac{1}{2}\sqrt{10}\,km{{s}^{-1}}\]
C) \[2.5\,\,km{{s}^{-1}}\]
D) \[\sqrt{\frac{15}{2}}\,\,km{{s}^{-1}}\]
Correct Answer: D
Solution :
The square root of \[{{\overline{v}}^{2}}\] is called the root mean square velocity (rms) speed of the molecules. \[{{v}_{rms}}=\sqrt{{{\overline{v}}^{2}}}\] \[=\sqrt{\frac{{{v}_{1}}^{2}+{{v}_{2}}^{2}+{{v}_{3}}^{2}+{{v}_{4}}^{2}}{4}}\] \[=\sqrt{\frac{{{(1)}^{2}}+{{(2)}^{2}}+{{(3)}^{2}}+{{(4)}^{2}}}{4}}\] \[=\sqrt{\frac{1+4+9+16}{4}}\] \[=\sqrt{\frac{30}{4}}\] \[=\sqrt{\frac{15}{2}}\,km{{s}^{-1}}\]You need to login to perform this action.
You will be redirected in
3 sec