InterviewSolution
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InterviewSolution
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`1mol` of an ideal gas undergoes reversible isothermal expansion form an initial volume `V_(1)` to a final volume `10V_(1)` and does `10kJ` of work. The initial pressure was `1xx 10^(7) Pa`. a. Calculate `V_(2)`. b. If there were `2mol` of gas, what must its temperature have been? |
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Answer» `W=-2.303 nRT"log"(V_(2))/(V_(1))` (i) Where `W` is work done by the system under isothermal reversible condition, note that work done by the system is negative `=-10xx10^(3)=-2.303xx1xx8.314xxT"log"(P_(1))/(P_(2))`....(1) Also `P_(1)V_(1)=P_(2)V_(2)` at constant temperature `1xx10^(7)xxV_(1)=P_(2)xx10V_(1)` `:. P_(2)=(1xx10^(7))/10=10^(6) Pa` `:.` By eq. (i) `-10xx10^(3)=-2.303xx1xx8.314xxT"log"(10^(7))/(10^(6))` `T=522.27 K` Now using, `PV=nRT` for `1` mole of gas, `P=1xx10^(7) Pa=10^(7) Nm^(-2)` `1xx10^(7)xxV_(1)=1xx8.314xx522.27` `V_(1)=4.34xx10^(-4) m^(3)` (ii) If `2` mole of gas have been used, the temperature would have been `522.27/2=261.13 K` |
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