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Standard XII

Chemistry

Maxwell's Distribution of Speeds

At 400 K, the...

Question

At 400K, the root mean square (rms) speed of a gas X (molecular weight=40) is equal to the most probable speed of gas Y at 60 K.The molecular weight of the gas Y is_________.

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Solution

Given that $u_{r m s} (X_{g a s}) a t 400 K = u_{m p} (Y_{g a s}) a t 60 K, m o l e c u l a r w e i g h t M_{1} = 40, M_{2} = ?, T_{1} = 400 K a n d T_{2} = 60 K .$

Therefore.

$\sqrt{\frac{3 R T_{1}}{M_{1}}} = \sqrt{\frac{2 R T_{2}}{M_{2}}}$

$\sqrt{\frac{400 \times 3 \times R}{40}} = \sqrt{\frac{2 \times R \times 60}{M_{2}}}$

Solving , we get $M_{2}$ = $4$ .

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At 400K, the root mean square (rms) speed of a gas X (molecular weight=40) is equal to the most probable speed of gas Y at 60 K.The molecular weight of the gas Y is_________.

Given that urms (Xgas) at 400K = ump(Ygas) at 60 K, molecular weight M1 = 40, M2= ? , T1 = 400K and T2= 60K. Therefore. √3RT1M1=√2RT2M2 √400 × 3 × R40 = √2 × R × 60M2 Solving , we get M2 = 4.