InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
`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`. c. Calculate `V_(2)`. b. If there were `2mol` of gas, what must its temperature have been? |
|
Answer» `w =- 2.303 nRt "log" (V_(2))/(V_(1))` where, `w` is work done by the system under isothermal reversible conditions, not that work done by the system is negative `-10 xx 10^(3) =- 2.303 xx 1xx 8.314 xx T "log"(P_(1))/(P_(2)) ……(i)` Also, `P_(1)V_(1) = P_(2)V_(2)` at constant temperature `1 xx 10^(7) xx V_(1) = P_(2) xx 10 V_(1)` `,. P_(2) = (1xx10^(7))/(10) = 10^(6) Pa` `:.` By equation (i), `- 10 xx 10^(3) =- 2.303 xx1xx 8.314 xxT "log" (10^(7))/(10^(6))` `:. T = 522.27K` Now, using `PV = nRT` for `1`mole of gas, `P = 1xx 10^(7) Pa = 10^(7) Nm^(-2)` `1 xx 10^(7) xx V_(1) = 1xx 8.314 xx 522.27` `V = 4.34 xx 10^(-4)m^(3)` b. If `2` mole of gas have been used, the temperature would have been `= (522.27)/(2) = 261.13K` |
|