The molecular kinetic energy of 1 g of helium (molecular weight 4) at `127^(@)C` is (Given , R `=8.31 J"mol"^(-1)K^(-1)` )

Question

The molecular kinetic energy of 1 g of helium (molecular weight 4) at `127^(@)C` is (Given , R `=8.31 J"mol"^(-1)K^(-1)` )

The molecular kinetic energy of 1 g of helium (molecular weight 4) at `127^(@)C` is (Given , R `=8.31 J"mol"^(-1)K^(-1)` )
A. 12.84 J
B. 12.465 J
C. 14.34
D. 14.384 J

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
Given , T = 273 + 127 = 400 K
`because` Helium is monoatmic gas.
`therefore` Average kinetic energy per mole of helium = `(3)/(2)RT`
Average kinetic energy of 1 g of helium = `(3)/(2)(RT)/(M)`
`=(3xx8.31xx4.00)/(2xx4)=12.465J`

4 views Answered 2023-02-05 15:29