partikel partikel gas oksigen didalam tabung tertutup pada suhu 20

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Persamaan energi kinetik gas:

E K equals 3 over 2 k T E K tilde T

Dari soal dikertahui

T subscript 1 equals 20 degree straight C plus 273 space straight K T subscript 1 equals 293 space straight K Ek subscript 1 equals 2140 space straight J Ek subscript 2 equals 6420 space straight J

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator E K subscript 1 over denominator E K subscript 2 end fraction end cell equals cell T subscript 1 over T subscript 2 end cell row cell fraction numerator 2140 space straight J over denominator 6420 space straight J end fraction end cell equals cell fraction numerator 293 space straight K over denominator T subscript 2 end fraction end cell row cell T subscript 2 end cell equals cell 293 cross times 6420 over 2140 straight K end cell row cell T subscript 2 end cell equals cell 879 space straight K end cell row cell T subscript 2 end cell equals cell left parenthesis 879 minus 273 right parenthesis space degree straight C end cell row cell straight T subscript 2 end cell equals cell 606 space degree straight C end cell end table

Dengan demikian, untuk menerima energi kinetik 6420 J suhunya sebesar 606 space degree straight C.

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