The average speed at T₁K and most probable speed at T₂K of CO₂ gas is $9\times {10}^4$ $\mathrm{cm}/\sec$. The values of ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$ are

Average speed: $\sqrt{\frac{8{\mathrm{RT}}_1}{{\mathrm{\pi M}}_1}}$

Most probable speed

For ${\mathrm{CO}}_2$
 

${\mathrm{V}}_{\mathrm{mp}}=\sqrt{\frac{2\mathrm{RT}}{\mathrm{M}}}$

Average speed at ${\mathrm{T}}_1=$Most probable speed at ${\mathrm{T}}_2=9\times {10}^4 \mathrm{cm}/\sec =9\times {10}^2$ $\mathrm{m}/\sec$.

Now,

$9\times {10}^2=\sqrt{\frac{8\times 8.314\times {\mathrm{T}}_1}{3.14\times 44\times {10}^{-3}}}\dots 1$

Most probable speed at T₂

$9\times {10}^2=\sqrt{\frac{8\times 8.314\times {\mathrm{T}}_2}{44\times {10}^{-3}}}\dots 2$

On solving equation 1 and 2 , we have

${\mathrm{T}}_1=1682.5 \mathrm{K}, {\mathrm{T}}_2=2143.4 \mathrm{K}$

Hence, correct option is (B)