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The average speed at $T_1K$ and the most probable speed at $T_2K of $CO_2$ gas is $9\times10^4cm/s. Calculate the value of $T_1 and T_2$

$\begin {array} {1 1}(a)\;T_1 = 1684.0K ; T_2 = 2143.37K\\(b)\;T_1=1234K;T_2 = 1298.3K\\(c)\;T_1=2378.3K;T_2=1369.3K\\(d)\;T_1=3456.1K;T_2=3139.6K \end {array}$

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Average speed = $\sqrt{(\large\frac{8RT}{\pi m})}$
Most probable speed = $\sqrt{(\large\frac{2RT}{m})}$
Average speed at $T_1 K$ = MP speed at $T_2K for\; CO_2$
$\sqrt{(\large\frac{8RT_1}{\pi m})} = \sqrt{(\large\frac{2RT_2}{m})}$
$\large\frac{T_1}{T_2} = \large\frac{\pi}{4}$------(i)
Also for $CO_2 u_{mp} = \sqrt{(\large\frac{2RT}{m})} = 9\times10^4$
$=\sqrt{\large\frac{2\times8.314\times10^7\times T_2}{44})}= 9\times10^4$
By eq (1) We get
$T_2 = 2143.37K$
$T_1 = 1684.0K$
Hence answer is (A)
answered Apr 5, 2014 by sharmaaparna1
 

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