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0.040g of He is kept in a closed contain...

`0.040g` of He is kept in a closed container initially at `100.0^(@)C.`The container is now heated. Neglecting the expansion of the container, Calculate the temperature at which the internal energy is increased by `12J`.

Text Solution

Verified by Experts

`n_(1) = (0.040)/(4) = 0.01 " " f = 3`
`T = 100+273 = 373 K`
`12 = Delta U = (f)/(2)n R Delta T " " = (3)/(2) xx 0.01 xx R (T_(2)-373)`
`(24)/(3 xx 0.01 xx R) = T_(1)-373`
`T_(2) = 373+(24)/(3 xx 0.01 xx R) = 469 = 196^(@)C`
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Knowledge Check

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