A sample of air is saturated with benzene (vapor pressure \[ = 100{\text{ mm Hg at }}298{\text{ K}}\]) at \[298{\text{ K}}\], \[750{\text{ mm Hg}}\] pressure. It is isothermally compressed to one third of its initial volume, the final pressure of the system is:
A.\[2250{\text{ torr}}\]
B.\[2150{\text{ torr}}\]
C.\[2050{\text{ torr}}\]
D.\[1950{\text{ torr}}\]
Answer
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Hint: The above question is based on Boyle's law. Pressure and volume are inversely proportional to each other and their product is constant. Pressure of air can be calculated by subtracting vapor pressure from total pressure given.
Formula used:
\[{{\text{P}}_1}{{\text{V}}_1} = {{\text{P}}_2}{{\text{V}}_2}\]
where ${{\text{P}}_{\text{1}}}{\text{ and }}{{\text{P}}_{\text{2}}}$ are initial and final pressures and ${{\text{V}}_{\text{1}}}{\text{ and }}{{\text{V}}_{\text{2}}}$ are initial and final volumes.
Complete step by step answer:
The above question is based on Boyle’s law of gaseous mixture. It states that pressure applied to a gas is inversely proportional to its volume at constant temperature and number of moles.
Since in the question we have been given that the process is isothermal, that no change in temperature is occurring during the reaction and variation of pressure with the number of moles is given. Hence the given conditions fit with Boyle’s law.
The problem is total pressure is given to us instead of pressure of air. Air also contains benzene vapors, so total pressure is equal to the pressure by benzene vapor and pressure by air.
\[{\text{Total pressure }} = {\text{ vapor pressure of benzene }} + {\text{pressure of air}}\]
\[{\text{pressure of air}} = {\text{Total pressure }} - {\text{vapor pressure of benzene}}\]]
All the values are given to us; hence we will substitute the values:
\[{\text{pressure of air}} = {\text{Total pressure }} - {\text{vapor pressure of benzene}}\]
\[ \Rightarrow {\text{pressure of air}} = ({\text{750}} - 100){\text{ mm Hg}} = {\text{ 650mm Hg}}\]
Let the initial volume is V and we have calculated the initial pressure. It is given to us that the final volume reduces to one third so final volume will be \[\dfrac{{\text{V}}}{3}\].
Substituting the values in the given formula:
\[{\text{650 mm Hg}} \times {\text{V}} = \dfrac{{\text{V}}}{3} \times {{\text{P}}_2}\]
Rearranging the above equation, we will calculate the final pressure as:
\[{{\text{P}}_2} = 1950{\text{ mm Hg}}\]
But benzene is still present in air so we need to add its vapor pressure too. So the final pressure will be \[(1950 + 100{\text{ )mm Hg}} = 2050\,{\text{mm Hg}}\]
The correct option is C.
Note:
Boyle's law only obeys when the temperature and number of moles remain constant if any one of them changes then the equation for Boyle’s law will also change. The one which will change will move to the denominator on either side of the equation.
Formula used:
\[{{\text{P}}_1}{{\text{V}}_1} = {{\text{P}}_2}{{\text{V}}_2}\]
where ${{\text{P}}_{\text{1}}}{\text{ and }}{{\text{P}}_{\text{2}}}$ are initial and final pressures and ${{\text{V}}_{\text{1}}}{\text{ and }}{{\text{V}}_{\text{2}}}$ are initial and final volumes.
Complete step by step answer:
The above question is based on Boyle’s law of gaseous mixture. It states that pressure applied to a gas is inversely proportional to its volume at constant temperature and number of moles.
Since in the question we have been given that the process is isothermal, that no change in temperature is occurring during the reaction and variation of pressure with the number of moles is given. Hence the given conditions fit with Boyle’s law.
The problem is total pressure is given to us instead of pressure of air. Air also contains benzene vapors, so total pressure is equal to the pressure by benzene vapor and pressure by air.
\[{\text{Total pressure }} = {\text{ vapor pressure of benzene }} + {\text{pressure of air}}\]
\[{\text{pressure of air}} = {\text{Total pressure }} - {\text{vapor pressure of benzene}}\]]
All the values are given to us; hence we will substitute the values:
\[{\text{pressure of air}} = {\text{Total pressure }} - {\text{vapor pressure of benzene}}\]
\[ \Rightarrow {\text{pressure of air}} = ({\text{750}} - 100){\text{ mm Hg}} = {\text{ 650mm Hg}}\]
Let the initial volume is V and we have calculated the initial pressure. It is given to us that the final volume reduces to one third so final volume will be \[\dfrac{{\text{V}}}{3}\].
Substituting the values in the given formula:
\[{\text{650 mm Hg}} \times {\text{V}} = \dfrac{{\text{V}}}{3} \times {{\text{P}}_2}\]
Rearranging the above equation, we will calculate the final pressure as:
\[{{\text{P}}_2} = 1950{\text{ mm Hg}}\]
But benzene is still present in air so we need to add its vapor pressure too. So the final pressure will be \[(1950 + 100{\text{ )mm Hg}} = 2050\,{\text{mm Hg}}\]
The correct option is C.
Note:
Boyle's law only obeys when the temperature and number of moles remain constant if any one of them changes then the equation for Boyle’s law will also change. The one which will change will move to the denominator on either side of the equation.
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