`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

(a) `w = 2.303nRT log. (V_(2))/(V_(1))`
`10 xx 10^(3) J= 2.303 xx 1 xx 8.314 xx T xx log . ( 10V_(1))/( V_(1))`
or `T = 522.3 K`
For initial conditions, ` P_(1)V_(1) = n_(1)RT`
i.e., `( 10^(7)Pa) V_91) = 1 xx 8.314 xx 522.3`
or ` V_(1) = 4.342 xx 10^(-4) m^(3) = 4.342 xx 10^(2) cm^(3) =434 .2 cm^(3)`
Note. We cannot apply the formula `-x = P Delta V` because expansion is not against constant pressure.
(b) If there were 2 moles of the gas, applying
`P_(1)V_91) = n_(1)RT ` , we get
`( 10^(7) Pa) ( 4.342 xx 10^(-4)m^(3)) =2 xx 8.314 xx T`
or`T = 261.1 K`, i.e. half of the first value