Question

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

Answer 1

Hint: 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 \[{O_2}\, = \,32g/mol\]

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: \[P\,V = \,nRT\]
Where,
\[P\] = Pressure of the gas
\[V\]=Volume of the gas
\[n\]=Number of moles of the gas
\[R\]=Universal gas constant
\[T\]=Temperature of the gas
According to the given question:
Given weight of oxygen \[{O_2}\, = \,370g\]
Molecular weight of oxygen \[{O_2}\, = \,32g/mol\]
We know that,
\[number\,of\,moles\,(n)\, = \dfrac{{Given\,\,weight}}{{Molecular\,\,weight}}\]
\[number\,of\,moles\,of\,{O_2}\,(n)\, = \dfrac{{370}}{{32}}\,mol\]
According to ideal gas equation, \[P\,V = \,nRT\]
Here, pressure = \[30.0\,atm\]
Temperature = \[{25^o}C\, = \,25\, + \,273\,\, = \,298K\]
Universal Gas constant \[R = 0.0821\,\,L\,\,atm\,\,mo{l^{ - 1}}\,{K^{ - 1}}\]
Putting all the values in ideal gas equation, we get
\[30\,\, \times \,\,V\, = \,\dfrac{{370}}{{32}}\, \times \,0.0821\, \times \,298\]
\[\,V\, = \,\dfrac{{370\,\, \times \,0.0821\, \times \,298}}{{32\,\, \times 30}}\,\]
We get,
\[V = \,9.4322\,L\]
Now, find number of moles using ideal gas equation when temperature is \[{75^o}C\]
\[P\,V = \,nRT\]
Here, Pressure = \[1.0\,atm\]
Volume = \[V = \,9.4322\,L\]
Temperature = \[{75^o}C\, = \,75\, + \,273\,\, = \,348K\]
Substituting all the values,
\[1\,\, \times \,\,9.432\,\, = \,\,n\,\, \times \,\,0.0821\,\, \times \,\,348\]
\[n = \,\dfrac{{0.0821\,\, \times \,348}}{{9.432}}\]
\[n = \,0.33\,mol\]
Weight of oxygen \[{O_2}\,\]= \[0.330\,\,mol\,\, \times \,32\,g/mol\]
Weight of oxygen \[{O_2}\,\]= \[10.6\,g\]
Mass of the oxygen escaped = \[370 - 10.6\]
Hence, Mass of the oxygen escaped = \[359.4\,g\]

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.