Saved Bookmarks
| 1. |
56 g of N and 96 g of O are mixed in 2 litrs container at 300 k. calculate the partial pressure of each gas and total pressure of the mixeure |
Answer» Answer:Given Mass of N = 56 g Given Mass of O = 96 g Moles of N = 56/28 = 2 moles Moles of O = 96/32 = 3 moles Mole Fraction of N = 2/ ( 2 + 3) = 2/5 Mole Fraction of O = 3/5 PARTIAL PRESSURE of N, P = KH (X) P = KH ( 2/5) P(N) = 6.51 × 2 × 10^7 /5 = 2.604 mm Partial Pressure of O, P = KH ( 3/5) KH for O2 = 3.30 × 10^7 mm P(O) = 3.30× 3 × 10^7 mm/ 5 = 1.98 × 10^7 mm Total Pressure of the MIXTURE = P(N) + P(O) = ( 2.60 + 1.98 ) × 10^7 mm Total Pressure, P(T ) = 4.58 × 10^7 mm. |
|