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A gas expends from `2 L` to `6L` against a constant pressure of `0.5` atm on absobing `200 J` of heat. Calculate the change in internal energy. |
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Answer» `Delta U = q+w` `= 200 J-0.5 atm (62-22)` `= 200 J-2 L atm` `= 200 J - 2 xx 101. 3 J ("1 L atm = 101.3 J")` `= - 2.6 J` |
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