Saved Bookmarks
| 1. |
Three moles of the ideal gas are expanded isothermally from a volume of 300 `cm^(3)` to 2.5 L at 300 K against a pressure of 1.9 atm. Calculate the work done in L and and joules . |
|
Answer» Correct Answer - Work of expansion =W= - 4.18 L atm `=- 423. 4J` Number of moles of a gas = n= 3 mol Initial volume `=V_(1) = 300 cm^(3) = 0.3 L (or dm^(3))` Final volume `= V_(2) = 2.5 L` External pressure `= P_(ex) = 1.9 atm ` Temperature = T = 300 K Since the expansion takes place against constant pressure it is irreversible process and work is given by `W=- P_("ex") (V_(2)-V_(1))` =- 1.9 (2.5 -0.3) `=- 4.18 L atm ` Now , 1 L atm = 101 . 3 J `:. W= - 4.18 xx 101 .3 =- 423 .4 J` |
|