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Find the final value of a gram molecule of a gas after an isothermal expansion at `127^(@)C`, if the original volume is `400 c.c` Given amount of work done by a gram molecule of a gas during expansion is `2302.6 "joule", R= 8.3 "joule mole"^(-1)K^(-1)` |
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Answer» Correct Answer - `800 c.c` `V_(2)= ?, T_(1)= 127+273= 400K, V_(1)= 400 c.c` `W=2302.6J , R= 8.3 "joule mole"^(-1)K^(-1)` As `W= 2.3026 RT "log"_(10)((V_(2))/(V_(1)))` `2302.6= 2.3026xx8.3xx400 "log"_(10)((V_(2))/(V_(1)))` `"log"_(10)((V_(2))/(V_(1)))= (2302.6)/(2.3026xx8.3xx400)` `=0.3012= log_(10)2` `:. (V_(2))/(V_(1))= 2, V_(2)= 2V_(1)= 2xx400= 800 c.c` |
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