7 gm of nitrogen is present at 127- C and 16 gm of oxygen at 27- C- Calculate the ratio of kinetic energy of nitrogen and oxygen-A- 2- 3B- 4- 5C- 4- 3D- 5- 4

The correct option is A 2 : 3We know, Kinetic energy =3/2nRT For two gases, ratio of kinetic energies =(K.E)_1/(K.E)_2=n_1T_1/n_2 T_2=7/28× 400 ×32/16×1/300= ...

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Standard XII

Chemistry

Molecular Speeds - Vrms, Vavg, Vmps

7 gm of nitro...

Question

7 gm of nitrogen is present at $127^{\circ} C$ and 16 gm of oxygen at $27^{\circ} C$ . Calculate the ratio of kinetic energy of nitrogen and oxygen.

4 : 3

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2 : 3

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4 : 5

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5 : 4

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Solution

The correct option is B 2 : 3
We know,
Kinetic energy $= \frac{3}{2} n R T$

For two gases, ratio of kinetic energies

$= \frac{(K . E)_{1}}{(K . E)_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}} = \frac{7}{28} \times 400 \times \frac{32}{16} \times \frac{1}{300} = 2 : 3$

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7 gm of nitrogen is present at 127 ∘C and 16 gm of oxygen at 27 ∘C. Calculate the ratio of kinetic energy of nitrogen and oxygen.

The correct option is B 2 : 3We know, Kinetic energy =32nRT For two gases, ratio of kinetic energies =(K.E)1(K.E)2=n1T1n2T2=728×400×3216×1300=2:3

7 gm of nitrogen is present at $127^{\circ} C$ and 16 gm of oxygen at $27^{\circ} C$ . Calculate the ratio of kinetic energy of nitrogen and oxygen.

The correct option is B 2 : 3
We know,
Kinetic energy $= \frac{3}{2} n R T$

For two gases, ratio of kinetic energies

$= \frac{(K . E)_{1}}{(K . E)_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}} = \frac{7}{28} \times 400 \times \frac{32}{16} \times \frac{1}{300} = 2 : 3$