`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`.

Options

A. `126^@C`
B. `258^@C`
C. `365^@C`
D. `196^@C`

Answer

D

Explanation

Given,
` m = 0.040 g, `
` T= (100^@C), (M_He)= 0.04g `
` U = (3/2)(nRT) = (3/2) xx (m/MRT)`
` T = ?`
` Given, (3/2 xx m/M) xx R xx T + 12`
` = (3/2) xx (m/M) xx R xx T`
` rArr 1.5 xx 0.01 xx 8.3 xx 373 +12 `
` = 1.5 xx 0.01 xx 8.3 T`
` rArr 58.4385 = 1245 T `
` rArr T = (58.4385/ 0.1245)`
` = 469.3855K`
` = 196.3 ^@C ~~ 196^@C`