A vessel contains 1.6 g of dioxygen at sTP (273.15 K, 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate (i) volume of the new vessel. (ii) number of molecules of dioxygen.

A vessel contains 1.6 g of dioxygen at sTP (273.15 K, 1 atm pressure).

1.	A vessel contains 1.6 g of dioxygen at sTP (273.15 K, 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate (i) volume of the new vessel. (ii) number of molecules of dioxygen.
Answer» SOLUTION :(i) `"1.6 g O"_(2)=1.6//32 mol="1.05 mol,1 mol of O"_(2)" at STP has volume of = 22.4 L"` `therefore"0.05 mol of O"_(2)" at STP has volume "=22.4xx0.05=1.12 L` (Remember. 1 mol of gas = 22.4 L if STP conditions are 273.15 K and atm `"= 22.7 L if STP conditions are 273.15 K and 1 bar")` `"Now,"V_(1)=1.12L,""P_(1)="1 atm,"V_(2)=?,""P_(2)=(1)/(2)" atm = 0.5 atm"` As temperature REMAINS constant, APPLYING Boyle's law, `P_(1)V_(1)=P_(2)V_(2) or V_(2)=(P_(1)V_(1))/(P_(2))=("1 atm"xx1.12L)/("0.5 atm")=2.24 L` (ii) Number of MOLECULES in 1.6 g `O_(2)`, i.e., 0.05 mol `O_(2)=0.05xx6.022xx10^(23)=3.011xx10^(22)`.