1. Tardigrade
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  3. Chemistry
  4. At what temperature will the total K.E. of 0.3 mol of He be the same as the total K.E. of 0.4 mol of Ar at 400 K

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Q. At what temperature will the total K.E. of 0.3 mol of He be the same as the total K.E. of 0.4 mol of Ar at 400 K

States of Matter

Solution:

K.E. of n moles of a gas,
K.E. = $\frac{3}{2}$nRT
K.E, = $\frac{3}{2} n_1 \,RT_1$
$K.E_2 = \frac{3}{2} n_2 \, RT_2$
$K.E_1 = K.E_2$
$n_1T_1 = n_2T_2$ 0.3 $\times$ T , = 0.4 $\times$ 400
$T_1 = \frac{1600}{3} $ = 533 K