Correct Answer - D
Pressure of dry gas, P=nRT/V
`=0.0168xx(0.0821xx1000xx760" ml "mm)xx(298)/(428)`
730 mm
Pressure of moist gas=754 mm. Hence, pressure of water vapour =(754-730)mm=24 mm.
Alternatively, volume of `0.0168" mol"` of `O_(2)` at STP
`=0.0168xx22400=376.3" ml"`
Thus, `V_(1)=376.3" ml", P_(1)=760" mm", T_(1)=273" K"`
`V_(2)=428" ml",P_(2)=?,T_(2)=298" K"`.
Calculate `P_(2)`.