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

Question

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 (A) 400 K (B) 300 K (C) 346 K (D) 533 K

Tardigrade Admin · Accepted Answer

K.E. of n moles of a gas, K.E. = (3/2)nRT K.E, = (3/2) n1 RT1 K.E2 = (3/2) n2 RT2 K.E1 = K.E2 n1T1 = n2T2 0.3 × T , = 0.4 × 400 T1 = (1600/3) = 533 K

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

400 K

300 K

346 K

533 K

K.E. of n moles of a gas,
K.E. = $\frac{3}{2}$ nRT
K.E, = $\frac{3}{2} n_{1} R T_{1}$
$K . E_{2} = \frac{3}{2} n_{2} R T_{2}$
$K . E_{1} = K . E_{2}$
$n_{1} T_{1} = n_{2} T_{2}$ 0.3 $\times$ T , = 0.4 $\times$ 400
$T_{1} = \frac{1600}{3}$ = 533 K

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

400 K

300 K

346 K

533 K

K.E. of n moles of a gas, K.E. = 23​nRT K.E, = 23​n1​RT1​ K.E2​=23​n2​RT2​ K.E1​=K.E2​ n1​T1​=n2​T2​ 0.3 × T , = 0.4 × 400 T1​=31600​ = 533 K

K.E. of n moles of a gas,
K.E. = $\frac{3}{2}$ nRT
K.E, = $\frac{3}{2} n_{1} R T_{1}$
$K . E_{2} = \frac{3}{2} n_{2} R T_{2}$
$K . E_{1} = K . E_{2}$
$n_{1} T_{1} = n_{2} T_{2}$ 0.3 $\times$ T , = 0.4 $\times$ 400
$T_{1} = \frac{1600}{3}$ = 533 K