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Question: The average kinetic energy of a gas molecule at 27°C is \(6.21 \times 10^{- 21}J\). Its average kine...

The average kinetic energy of a gas molecule at 27°C is 6.21×10−21J6.21 \times 10^{- 21}J. Its average kinetic energy at 227°C will be

A

(a) 52.2×10−21J52.2 \times 10^{- 21}J

B

5.22×10−21J5.22 \times 10^{- 21}J

C

10.35×10−21J10.35 \times 10^{- 21}J

D

11.35×10−21J11.35 \times 10^{- 21}J

Answer

10.35×10−21J10.35 \times 10^{- 21}J

Explanation

Solution

E∝TE \propto T̃E1E2=T1T2\frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}}̃ 6.21×10−21E2=(273+27)(273+227)=300500\frac{6.21 \times 10^{- 21}}{E_{2}} = \frac{(273 + 27)}{(273 + 227)} = \frac{300}{500}

̃ E2=10.35×10−21JE_{2} = 10.35 \times 10^{- 21}J