Calculate no of degrees of freedom in 10 cm3 of o2
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Explanation: 1 molecule of o2 at ntp =5 D.O.F
22400cm3= 6.022×10 power 23 molicule
10cm3 =6.022×10 power 23×10÷ 22400 molicules
There fore Dof= 6.022×10 power 23 ×10×5÷22400= 134.41×10 power 19
= 1.34×10power21
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