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Q.

The average and total kinetic energies of 0.5 mol  of  an ideal gas at   273 K are respectively in  KJ mot":

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a

2.5×102;3.9

b

3.405;1.7025

c

5.0×104:7.8

d

6.81;4.08

answer is B.

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Detailed Solution

R=8.314JK−mol−;T=273K. :                                  we know:(a) Average    K.E. =32RT=32                                   8.314JK−mol−×273K                    =3404.5Jmol−=3404.51000≅3.405kJmol− Total K.E. of   0.5 mol of gas                 =0.5mol×3.405kJmol−=1.7025kJ.
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The average and total kinetic energies of 0.5 mol  of  an ideal gas at   273 K are respectively in  KJ mot":