The ideal gas temperature dropped from t1 = 567 ° C to t2 = 147 ° C. In this case, the average kinetic

The ideal gas temperature dropped from t1 = 567 ° C to t2 = 147 ° C. In this case, the average kinetic energy of motion of gas molecules …

The solution to this problem:
E (kin) = 3/2 K * T
Let’s translate into kelvin: 567 + 273 = 840; 147 + 273 = 420
Answer: the temperature has halved



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