At what temperature, the translational kinetic energy of 14 g of nitrogen will be the same as that of 32g of oxygen at 300 K

150 K

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300 K

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600 K

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131. 25 K

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Solution

The correct option is C
600 K

$K E = \frac{3}{2} n R T; n_{N_{2}} = \frac{14}{28} = 0.5 m o l; T_{N_{2}} = ? n_{O_{2}} = \frac{32}{32} = 1 m o l; T_{O_{2}} = 300 K G i v e n, K . E . (N_{2}) = K . E . (O_{2}), S o n_{N_{2}} T_{N_{2}} = n_{O_{2}} T_{O_{2}}$
Or $0.5 \times T_{N_{2}} = 1 \times 300, o r T_{N_{2}} = 600 K$

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