Question

At _________ atm of pressure the volume of gas at 5 atm will be doubled at constant temperature.

Answer 1

Hint: The above question is based upon Boyle's law for gases. We need to know the definition of Boyle's law and putting initial value of pressure and the given ratio of initial volume and final volume, we will get the value of final pressure.

Complete step by step solution:
Boyle’s law states that the pressure of gas is inversely proportional to the volume of had provided that temperature and number of moles remain constant. Inversely proportional means that when pressure increases, volume will decrease and when pressure will increase then temperature will decrease. Let us look at the statement numerically,
\[{\text{P}} \propto \dfrac{1}{{\text{V}}}\] Where P is pressure and V is volume. Let K be any constant of proportionality, hence we will write the equation as:
\[{\text{P}} = \dfrac{{\text{K}}}{{\text{V}}}\]. Rearranging the above equation we will get
\[{\text{PV}} = {\text{K}}\]
Let \[{{\text{P}}_1}{\text{ and }}{{\text{V}}_1}\] be the initial pressure and initial volume respectively and \[{{\text{P}}_2}{\text{ and }}{{\text{V}}_2}\] be the final pressure and final volume respectively.
According to boyle’s law
\[{{\text{P}}_1}{{\text{V}}_1} = {\text{K}}\]
\[{{\text{P}}_2}{{\text{V}}_2} = {\text{K}}\]
Hence it is clear that, \[{{\text{P}}_1}{{\text{V}}_1} = {{\text{P}}_2}{{\text{V}}_2}\]
Now it is given to us in the question that initial pressure is 5 atm that is \[{{\text{P}}_1} = 5{\text{ atm}}\]and final volume is double the initial volume that is \[{{\text{V}}_2} = 2{{\text{V}}_1}\]. Putting the values in the equation:
\[5{\text{ }} \times {{\text{V}}_1} = {{\text{P}}_2} \times {\text{2}}{{\text{V}}_1}\]
Solving the above equations we will get \[\dfrac{{5{\text{ }} \times {{\text{V}}_1}}}{{{\text{2}}{{\text{V}}_1}}} = {{\text{P}}_2}\]
\[{{\text{P}}_2} = 2.5{\text{ atm}}\]

Note: The equation we used for boyle’s law that is \[{{\text{P}}_1}{{\text{V}}_1} = {{\text{P}}_2}{{\text{V}}_2}\] is only valid when temperature is constant for given set of equation. In any case the temperature varies the equation will be \[\dfrac{{{{\text{P}}_1}{{\text{V}}_1}}}{{{{\text{T}}_1}}} = \dfrac{{{{\text{P}}_2}{{\text{V}}_2}}}{{{{\text{T}}_2}}}\]. Sometimes even number of moles and temperature both varies in that case the equation will be:
 \[\dfrac{{{{\text{P}}_1}{{\text{V}}_1}}}{{{{\text{n}}_1}{{\text{T}}_1}}} = \dfrac{{{{\text{P}}_2}{{\text{V}}_2}}}{{{{\text{n}}_2}{{\text{T}}_2}}}\]. If only number of moles varies than the equation will be:
 \[\dfrac{{{{\text{P}}_1}{{\text{V}}_1}}}{{{{\text{n}}_1}}} = \dfrac{{{{\text{P}}_2}{{\text{V}}_2}}}{{{{\text{n}}_2}}}\].