`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 internal energy is increased by `12J`. `[R = (25)/(3)J-mol^(-1)K^(-1)]`
Text Solution
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`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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