A quantity of an ideal gas is collected in a graduated tube over the mercury. The volume of gas at 20.C is 50 ml and the level of the mercury is 100mm above the outside of the mercury level. The atmospheric pressure is 750mm. Volume of gas at STP is? (Take R = 0.083 litre atm/mole )

P1 = 650 torrV1 = 50 mlT1 = 20 C + 273 = 293 KP2 = 760 TorrV2 = ?T2 = 273 Kideal gas law:P1 V1 / T1 = P2 V2 / T2rearanges to:V2 = ( P1 V1 T2 ) / ( P2 T1 )V2 = (650) (50) (273) / (760) (293)V2 = 39.84 mlrounding off, much like your "50 ml" was rounded offwould give youV2 = 40 ml

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the level of mercury in the tube been 100 mm BELOW the outside of the mercury level, the gas would have been under more that atmospheric pressure: 750 mm Hg of air pressure + 100 mm of Mercury pressure but since the level of mercury in the tube 100 mm above the outside of the mercury level, the gas would have been under Less that atmospheric pressure: 750 mm Hg of air pressure - 100 mm of Mercury pressure = 650 mm Hg P1 = 650 torr V1 = 50 ml T1 = 20 C + 273 = 293 K P2 = 760 Torr V2 = ? T2 = 273 K ideal gas law: P1 V1 / T1 = P2 V2 / T2 rearanges to: V2 = ( P1 V1 T2 ) / ( P2 T1 ) V2 = (650) (50) (273) / (760) (293) V2 = 39.84 ml rounding off, much like your "50 ml" was rounde Thumbs up
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