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.