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Equipartition Theorem

calculate the...

Question

calculate the kinetic energy of a gram molecule of oxygen at 127 C kb=1,381 10^-23 n=6.02210^23

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Solution

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$K i n e t i c e n e r g y o f o n e m o l e o f a n i d e a l g a s p e r d e g r e e o f f r e e d o m = \frac{1}{2} K_{B} T \times N_{A} K i n e t i c e n e r g y o f a n i d e a l g a s f o r f d e g r e e o f f r e e d o m = \frac{1}{2} K_{B} T \times N_{A} \times f d e g r e e o f f r e e d o m o f d i a t o m i c g a s f = 5 k i n e t i c e n e r g y o f o n e m o l e o f a d i a t o m i c g a s K = \frac{1}{2} K_{B} T \times N_{A} \times 5 K = \frac{1}{2} \times 1.381 \times 10^{- 23} \times 400 \times 6.022 \times 10^{23} \times 5 = 8316.38 j o u l e R e g a r d s$

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6.02 \times 10^{23}

1.38 \times 10^{- 23}

27^{\circ}

27^{\circ}

27^{\circ}

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127^{\circ} C

R = 8320 J / m o l K

= 40 k g / m o l

6.022 \times 10^{23}

6.022 \times 10^{23}

6.022 \times 10^{23}

C O_{2}

6.022 \times 10^{23}

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H = 1, C = 12, N = 14, O = 16

One mole of oxygen gas at STP is equal to...........

(a) $6.022 \times 10^{23}$ molecules of oxygen

(b) $6.022 \times 10^{23}$ atoms of oxygen

(d) 32 g of oxygen

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