a veseel contains 1 6g of dioxygen at stp the gas is now transferd to anther vessel at constant temparature where pressure becomes half of the original pressure caluclate a)volume of new vessel b)numberer of molecules of dioxygen

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Solution

(a) The given data are at STP p1=1 atm, T1=273 K, V1=?

32 g of oxygen occupies 22.4 L of volume at STP*

Hence, 1.6 g of oxygen will occupy, 1 .6 g oxygen × 22.4 L/32 g oxygen = 1.12 L
V1=1.12 L

p2 = p1/2 = 1/2 = 0 .5 atm.

V2=?
According to Boyle’s law :
p1V1 = p2V2
V2 = p1 x V1 / p2 = 1 atm. × 1 .12 L / 0 .5 a tm. = 2.24 L

* Old STP conditions 273.15 K, 1 atm, volume occupied by 1 mol of gas = 22.4 L.
New STP conditions 273.15 K, 1 bar, volume occupied by a gas = 22.7 L.

(b) Number of molecules of oxygen in the vessel = 6 .022 x 10^23 x 1.6 / 32
= 3.011 × 10^22

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