Question

10) The average speed at T K and most probable
speed at T2K of CO, gas is 9 x 104 cm/sec.
The values of T, and T2 are
1) 2143 K; 1694 K 2) 1726 K;2126 K
3) 1684 K ; 2143 K 4) 1684 K ;3368K



plz solve it correctly ​

Answer 1

Answer:

answer is (1)

Explanation: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 (1)

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Answer 2

Answer:

answer is (1)

Explanation: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 (1)