`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?

Explanation

`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`