Question

What volume of oxygen at NTP is necessary to complete combustion of 20 litres of propane measured at ${{27}^{o}}C$ and 760 mm of Hg pressure?

Answer 1

Hint: A modification of the ideal gas law can be used to compare and find the volume of oxygen required at NTP. NTP means that temperature is ${{20}^{o}}C$ and pressure is 1 atm (760 mm of Hg).
Formula used: \[{{P}_{1}}{{V}_{1}}/{{T}_{1}}={{P}_{2}}{{V}_{2}}/{{T}_{2}}\]
  Complete answer:
- We first need to understand what is the reaction happening when combustion of propane occurs.
- Combustion of propane leads to the formation of carbon dioxide and water.
- The balanced chemical equation for this reaction is –
     \[{{C}_{3}}{{H}_{8}}+5{{O}_{2}}\to 3C{{O}_{2}}+4{{H}_{2}}O\]
- We have to first find the amount of oxygen necessary to complete combustion of 20 litres of propane measured at ${{27}^{o}}C$ and 760 mm of Hg pressure.
- We can easily understand from the balanced equation that 1 mol of \[{{C}_{3}}{{H}_{8}}\] requires 5 moles of \[{{O}_{2}}\] for combustion.
- The volume of propane is measured to be 20 litres.
- Hence, the volume of oxygen needed at ${{27}^{o}}C$ and 760 mm of Hg pressure for combustion of propane will be – 20 x 5 = 100 litres.
- Now that we have found the volume of oxygen needed at ${{27}^{o}}C$ and 760 mm of Hg, we need to find the volume of oxygen needed at NTP.
- Normal temperature and pressure assumes a temperature of ${{20}^{o}}C$ and a pressure of 1 atm (760 mm of Hg).
- Hence, to find the volume of oxygen needed at NTP, we can use the modified form of ideal gas law-

\[{{P}_{1}}{{V}_{1}}/{{T}_{1}}={{P}_{2}}{{V}_{2}}/{{T}_{2}}\]
- Let us consider that the conditions at NTP are \[{{P}_{1}}=1atm=760mmHg\], \[{{T}_{1}}={{20}^{0}}C=293K\].
- We already know that, \[{{P}_{2}}=1atm=760mmHg\], \[{{T}_{2}}={{27}^{0}}C=300K\],\[{{V}_{2}}=100L\].

- Substituting these values in the modified ideal gas equation, we can find the volume of oxygen needed as –
\[{{V}_{1}}=({{P}_{2}}{{V}_{2}}/{{T}_{2}})\times ({{T}_{1}}/{{P}_{1}})=(1\times 100/300)\times (293/1)=293/3=97.667L\]
 - Hence, we can conclude that 99.667 L of oxygen at NTP is needed for complete combustion of 20 litres of propane measured at ${{27}^{o}}C$ and 760 mm of Hg pressure.

Note: It is very important that you measure the amount of oxygen required at NTP and not at ${{27}^{o}}C$ and 760 mm of Hg pressure. The answer for this question is not 100 litres. 100 litres is the volume of oxygen needed at ${{27}^{o}}C$ and 760 mm of Hg pressure for combustion of propane.