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Question: A gas cylinder contains \[370\,g\] oxygen at \[30.0\,atm\] pressure and \[{25^o}C\]. What mass of ox...

A gas cylinder contains 370g370\,g oxygen at 30.0atm30.0\,atm pressure and 25oC{25^o}C. What mass of oxygen will escape if the cylinder is first heated as 75oC{75^o}C and then the valve is held open until gas pressure becomes 1.0atm1.0\,atm , the temperature being maintained at 75oC{75^o}C?

Explanation

Solution

Here, we can use the ideal gas equation to solve this question. It is an empirical relation among the volume, the temperature, the pressure, and the number of moles of the gas. It is an expression which describes the behavior of gas. Take Molecular weight of oxygen O2=32g/mol{O_2}\, = \,32g/mol

Complete answer:
An ideal gas is a hypothetical gas whose behavior is completely independent of attractive and repulsive forces and defined by the ideal gas law. The ideal gas relation is the empirical relation among the volume, the temperature, the pressure, and the number of moles of the gas. The ideal gas equation is used to calculate the value of the any quantity. It is also used to predict the value when the conditions are changed and if we know the original conditions.
Ideal gas equation: PV=nRTP\,V = \,nRT
Where,
PP = Pressure of the gas
VV=Volume of the gas
nn=Number of moles of the gas
RR=Universal gas constant
TT=Temperature of the gas
According to the given question:
Given weight of oxygen O2=370g{O_2}\, = \,370g
Molecular weight of oxygen O2=32g/mol{O_2}\, = \,32g/mol
We know that,
numberofmoles(n)=GivenweightMolecularweightnumber\,of\,moles\,(n)\, = \dfrac{{Given\,\,weight}}{{Molecular\,\,weight}}
numberofmolesofO2(n)=37032molnumber\,of\,moles\,of\,{O_2}\,(n)\, = \dfrac{{370}}{{32}}\,mol
According to ideal gas equation, PV=nRTP\,V = \,nRT
Here, pressure = 30.0atm30.0\,atm
Temperature = 25oC=25+273=298K{25^o}C\, = \,25\, + \,273\,\, = \,298K
Universal Gas constant R=0.0821Latmmol1K1R = 0.0821\,\,L\,\,atm\,\,mo{l^{ - 1}}\,{K^{ - 1}}
Putting all the values in ideal gas equation, we get
30×V=37032×0.0821×29830\,\, \times \,\,V\, = \,\dfrac{{370}}{{32}}\, \times \,0.0821\, \times \,298
V=370×0.0821×29832×30\,V\, = \,\dfrac{{370\,\, \times \,0.0821\, \times \,298}}{{32\,\, \times 30}}\,
We get,
V=9.4322LV = \,9.4322\,L
Now, find number of moles using ideal gas equation when temperature is 75oC{75^o}C
PV=nRTP\,V = \,nRT
Here, Pressure = 1.0atm1.0\,atm
Volume = V=9.4322LV = \,9.4322\,L
Temperature = 75oC=75+273=348K{75^o}C\, = \,75\, + \,273\,\, = \,348K
Substituting all the values,
1×9.432=n×0.0821×3481\,\, \times \,\,9.432\,\, = \,\,n\,\, \times \,\,0.0821\,\, \times \,\,348
n=0.0821×3489.432n = \,\dfrac{{0.0821\,\, \times \,348}}{{9.432}}
n=0.33moln = \,0.33\,mol
Weight of oxygen O2{O_2}\,= 0.330mol×32g/mol0.330\,\,mol\,\, \times \,32\,g/mol
Weight of oxygen O2{O_2}\,= 10.6g10.6\,g
Mass of the oxygen escaped = 37010.6370 - 10.6
Hence, Mass of the oxygen escaped = 359.4g359.4\,g

Note:
It should be noted that ideal gas is a hypothetical gaseous substance, in reality there is no such ideal gas. It acts as a useful equation and helps in understanding how gases respond when conditions are changed.