Question

At which temperature n2 molecule have same average speed as he atoms at 330k

Answer 1

Hey dear,

◆ Answer -
2310 K

◆ Explaination -
# Given-
T1 = 330 K
T2 = ?
M1(for He) = 4 g
M2(for N2) = 28 g

# Solution-
Given situation is
vavg(He) = vavg(N2)
√(3n1RT1/2) = √(3n2RT2/2)
√(T1/M1) = √(T2/M2)
T2 = T1 × M2 / M1
T2 = 330 × 28 / 4
T2 = 2310 K

At 2310 K N2 will have same avg speed to that of He at 330 K.

Hope this helps...