What is the kinetic energy of $1$ g of $O_{2}$ at $47^{o}$ C?

1.24 \times 10^{2} J

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2.24 \times 10^{2} J

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1.24 \times 10^{3} J

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3.24 \times 10^{2} J

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Solution

The correct option is A $1.24 \times 10^{2} J$
Moles of $O_{2} = \frac{1}{32} m o l e s$

K. E. of n moles $= \frac{3 n R T}{2}$

K. E. of $\frac{1}{32}$ moles $= \frac{3 \times 1 \times 8.314 \times 320}{2 \times 32}$

$= 1.24 \times 10^{2} J$

Hence, option A is correct.

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What is the kinetic energy of 1g of O2 at 47oC?

The correct option is A 1.24×102JMoles of O2=132 molesK. E. of n moles =3nRT2K. E. of 132 moles =3×1×8.314×3202×32 =1.24×102 JHence, option A is correct.

What is the kinetic energy of $1$ g of $O_{2}$ at $47^{o}$ C?

The correct option is A $1.24 \times 10^{2} J$
Moles of $O_{2} = \frac{1}{32} m o l e s$

K. E. of n moles $= \frac{3 n R T}{2}$

K. E. of $\frac{1}{32}$ moles $= \frac{3 \times 1 \times 8.314 \times 320}{2 \times 32}$

$= 1.24 \times 10^{2} J$

Hence, option A is correct.