Saved Bookmarks
| 1. |
1.22 g of a gas measured over water at `150^(@)C` and a pressure of 775 mm of mercury occupied 900 mL. Calculate the volume of dry gas at NTP. Vapour pressure of water at `15^(@)C` is 14 mm. |
|
Answer» Pressure of dry gas = Pressure of moist gas `-` Aqueous tension `= 775 - 14` `= 761 mm` `{:("Initial conditions","NTP Conditions",),(V_(1) = 900 mL,V_(2) = ?,),(P_(1) = 761 mm,P_(2) = 760 mm,),(T_(1) = (273 + 15) = 288 K,T_(2) = 273 K,):}` Since, `(P_(1)V_(1))/(T_(1)) = (P_(2) V_(2))/(T_(2))` So, `V_(2) = (P_(1) V_(1) T_(2))/(T_(1) P_(2))` `= (761 xx 900 xx 273)/(288 xx 760)` `= 854.2 mL` |
|