Question

09 A gas with a volume of 525 mL at a temperature of -25 °C is heated to 175 °C. What is the new volume, in milliliters, of the gas if pressure and number of moles are held constant? 010 A gas has a volume of 2.8 L at a temperature of 27 °C. What temperature (°C) is needed to expand the volume to 15 L? (P and n are constant.) a 1971 011 Combined gas law problem: A balloon is filled with 500.0 mL of helium at a temperature of 27 °C and 755 mmHg. What volume, in milliliters, will it have when it reaches an altitude where the temperature is -33 °C and the pressure is 0.65 atm? 500 mL = V 755 mmHg =P - 33 CT2 0.65 atm =P 27°C = T 10

Answer 1

The concept used in question 9 and question 10 is that at constant pressure and number of moles of gas the volume of gas is directly proportional to temperature of gas and thus charle's law can be applied to get final volume or temperature. In question 10 the final volume can be found using combined gas law as mentioned in question. The one thing that has to be remembered is that units taken in formula of these physical properties should be same

Date Question - Given Initial volume (w) of gas = 525 ml Initial Temperature (1 of gas = -25°C - or 240K Final Temperature (T2) of gas - = 75oC ox 448 k Final volume ( var Pressure and no. of moles are constant Charle's law can be applied So temp. must be in Kelvin - v = 448x525 948.387 ml 248 final volume (V₂) = 948.387 m .

Question - 10 Given Initial volume lvl of gas = 2.8L Initial timp. (TJ of gas = 27° C or 3ook -Final volume of gase) = 15L 2 Final temperature of the gos (I) = since; p and n Charle's are law constant can be applied N - ₂ TL Tool toge 2. 8L - ISL 300 2 T = 2.8k T2 = 158300 - 4607.142k 2.8 final temp. Ia in °C = 1607.142 - 273. = 1334.14 °C

Question No.lt Initial volume of Helium gas (4) = 500 ml Initial temperature of gas = 27°C or 300k Initial Pressure of gas (P.) = 755mm Hg final temperature of gas (T) = -339 or 240K Final pressure of gas (P) = 0.65 atm or 0.658760 mm Ho = 494 mm Hg I atm = 760 mm Hal - Imp. point thits of thisted 1 Units on both side of should be same. combined gas law, formula Using Pivah V - 755 mm Hg x 500 m) 300 k 4gu mm Hg x 4 240k Voz 494x300 7:5XSDO X240 494 x 300 - V: 611.336 ml final vol of gas (1) = 611.336 ml.