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Let one mole of a  substance of atomic radius r and density  have molar mass M Let us assume the atoms to be spherical Avogadro's number is  For Carbon For gold For Nitrogen For Lithium For Fluorine
As per Graham's Law of diffusion if two gases of Molar Mass M1 and M2 diffuse with rates R1 and R2 respectively their diffusion rates are related by the following equation In the given question R1 = 28.7 cm3 s-1 R2 = 7.2 cm3 s-1  M1 = 2 g   The above Molar Mass is close to 32, therefore, the gas is Oxygen.  
Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P1 = 1 atm = 76 cm of Mercury Let the crossectional area of the tube be x cm2 The initial volume of the air column, V1 = 15x cm3 Let's assume once the tube is held vertical y cm of Mercury flows out of it. The pressure of the air column after y cm of Mercury has flown out of the column P2 = 76 - (76 - y)...
Pressure, P = 2atm Temperature, T = 17 oC The radius of the Nitrogen molecule , The molecular mass of  N2 = 28 u The molar mass of N2 = 28 g From ideal gas equation The above tells us about the number of moles per unit volume, the number of molecules per unit volume would be given as The mean free path  is given as  The root mean square velocity vrms is given as  The time between...
As we know root mean square velocity is given as  Let at temperature T the root mean square speed of an atom in an argon cylinder equal to the rms speed of a helium gas atom at
As per Avogadro's Hypothesis under similar conditions of temperature and pressure equal volumes of gases contain equal number of molecules. Since the volume of the vessels are the same and all vessels are kept at the same conditions of pressure and temperature they would contain equal number of molecules. Root mean square velocity is given as As we can see vrms is inversely proportional to the...
The volume of the room, V = 25.0 m3 Temperature of the room, T = 27oC = 300 K The pressure inside the room, P  = 1 atm Let the number of moles of air molecules inside the room be n Avogadro's Number,  Number of molecules inside the room is N
Initial Volume of the bubble, V1 = 1.0 cm3 Initial temperature, T1 = 12 oC = 273 + 12 = 285 K Density of water is  Initial Pressure is P1 Depth of the bottom of the lake = 40 m Final Temperature, T2 = 35 oC = 35 + 273 = 308 K Final Pressure = Atmospheric Pressure  Let the final volume be V2 As the number of moles inside the bubble remains constant we have
Initial volume, V1 = Volume of Cylinder = 30 l Initial Pressure P1 = 15 atm Initial Temperature T1 = 27 oC = 300 K Initial number of moles n1 inside the cylinder is Final volume, V2 = Volume of Cylinder = 30 l Final Pressure P2 = 11 atm Final Temperature T2 = 17 oC = 290 K Final number of moles n2 inside the cylinder is Moles of oxygen taken out of the cylinder = n2 -n1 = 18.28 - 13.86 =...
(a) The dotted plot corresponds to the ideal gas behaviour. (b) We know the behaviour of a real gas tends close to that of ideal gas as its temperature increases and since the plot corresponding to temperature T1 is closer to the horizontal line that the one corresponding to T2 we conclude T1 is greater than T2. (c)  As per the ideal gas equation The molar mass of oxygen = 32 g R = 8.314 (d)...
Diameter of an oxygen molecule, d = 3 Å. The actual volume of a mole of oxygen molecules Vactual is The volume occupied by a mole of oxygen gas at STP is Vmolar = 22.4 litres
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