InterviewSolution
Saved Bookmarks
| 1. |
The work done during the expanision of a gas from a volume of `4 dm^(3)` to `6 dm^(3)` against a constant external pressure of 3 atm is (1 L atm = 101.32 J)A. `-6 J`B. `-608 J`C. `+304 J`D. `-304 J` |
|
Answer» Correct Answer - B Work done (W) `=-P_("ext") (V_(2)-V_(1))` `=-3xx(6-4)=-6 L atm` `=-6xx101.32` J `(because` 1 L atm =101.32 J`)` `=-607.92 ~~-608 J` |
|