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larisa86 [58]
3 years ago
7

What is the temperature of a sample of gas when the average translational kinetic energy of a molecule in the sample is 8.37 × 1

0 − 21 J ?
Physics
1 answer:
-BARSIC- [3]3 years ago
4 0

Answer:

404K

Explanation:

Data given, Kinetic Energy.K.E=8.37*10^-21J

Note: as the temperature of a is increase, the rate of random movement will increase, hence leading to more collision per unit time. Hence we can say that the relationship between the kinetic energy and the temperature is a direct variation.

This relationship can be expressed as

K.E=\frac{3}{2}KT

where K is a constant of value 1.38*10^-23

Hence if we substitute the values, we arrive at

T=\frac{2/3(8.37*10^{-21})}{1.38*10^-23}\\ T=404K

converting to degree we have 131^{0}C

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