Question

If $250mL$ of ${{N}_{2}}$ ​ over water at $30{}^\circ C~$ and a total pressure of $740torr$ is mixed with $300mL$ of $Ne$ over water at $25{}^\circ C~$ and a total pressure of $780torr,$ what will be the total pressure if the mixture is in a $500mL$ vessel over water at $35{}^\circ C~.$
(Given : Vapour pressure (Aqueous tension) of ${{H}_{2}}O~$ at $25{}^\circ C~$ , $30{}^\circ C~$ and $35{}^\circ C~$ are $23.8,31.8$ and $42.2torr~$ respectively. Assume volume of ${{H}_{2}}{{O}_{(l)~}}$ is negligible in final vessel)
(A) $760torr$
(B) $828.4torr$
(C) $807.6torr$
(D) $870.6torr$

Answer 1

Hint : Use the equation of ideal gas law. Use Boyle’s law equation to and therefore the relation between pressure and volume. Apply the given conditions to the best gas law equation and what happens to the n value. Since all the parameters except the pressure are given within the question, we tend to use the equation of ideal gas to and out the desired pressure.

Complete step by step solution:
In order to answer our question, we'd like to use the idea of the ideal gas equation. Ideal gas equation is an equation that is followed by the best gases. A gas that might adapt Boyle's and Charles Law beneath all the conditions of pressure and temperature is termed a perfect gas. An ideal gas equation is shown as:
$PV=nRT$
This is the ideal gas equation because it is obeyed by the hypothetico gases known as ideal gases beneath all conditions of temperature and pressure but there's no gas that's perfectly ideal. However the gases could show nearly ideal behaviour beneath the conditions of low pressure and hot temperature and are known as real gases.
Firstly we will find number of moles for ${{N}_{2}}$
${{n}_{{{N}_{2}}}}=\dfrac{\left( \dfrac{708.2}{760}\times 0.25 \right)}{0.0821\times 303}$
${{n}_{{{N}_{2}}}}=9.36\times {{10}^{-3}}$
Firstly we will find number of moles for ${{O}_{2}}$
${{n}_{{{N}_{2}}}}=\dfrac{\left( \dfrac{756.2}{760}\times 0.3 \right)}{0.0821\times 298}$
${{n}_{{{N}_{2}}}}=0.012$
Therefore number of total moles is given by, ${{n}_{total}}\text{ }moles=0.0215$
Thus the final pressure ${{P}_{f}}$ in the vessel is given by ideal gas equation;
${{P}_{f}}=\dfrac{\left( {{n}_{total}} \right)RT}{V}$
Here ${{P}_{f}}$ is final pressure in vessel i.e. for ${{P}_{f}}={{P}_{{{N}_{2}}}}+{{P}_{{{O}_{2}}}}$
Substituting all the values in above equation to find ${{P}_{f}}$
${{P}_{f}}=\dfrac{0.0215\times 0.082\times 308}{0.5}$
${{P}_{f}}=1.09~atm~$ or ${{P}_{f}}=~828.4~~torr$
Therefore, total pressure is given by, $Ptotal=\left( {{P}_{{{N}_{2}}}}+{{P}_{{{O}_{2}}}} \right)+Vapour~pressure~of~{{H}_{2}}O$
$P_{total}=828.4+42.2$
$P_{total}=870.6torr$
Therefore, the correct answer is Option D i.e. the total pressure if the mixture is in a $500ml$ vessel over water at $35{}^\circ C$ is $870.6torr.$

Note:
Note that the Boyle’s law and Charles law can be combine to give relationship between variables $P,V$ and $T$ which is known as combine gases equation, and given by; $\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}$