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At what temperature the kinetic energy of a gas molecule is one-half of its value at 30^@C? |
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Answer» Solution :`(K.E.) =3/2 KT` At TWO different conditions ` ((KE)_1)/((KE)_2) = (T_1) /(T_2) , (1)/( 1//2)= ( 303)/(T_2)` Temperature at which KINETIC energy is half =`(303)/(2)` ` = 151.5 K =- 121.5^@ C ` |
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