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Gas molecules each of mass `10^(-26)` kg are taken in a container of volume `1 dm^(3)` . The root mean square speed of gas molecules is 1 km `sec^(-1)` .What is the temperature fo gas molecules. (Given : `N_(A) =6xx10^(23),R=8J//mol.K`)A. 298 KB. 25 KC. 250 KD. 2500 K |
Answer» Correct Answer - c (c) `1000=sqrt((3RT)/(M)),10^(3)=[(3xx8.314xxT)/(10^(-26)xx6xx10^(23))]^(1//2)` ` T=(10^(6)xx10^(-26)xx6xx10^(23))/(3xx8)=250K` |
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