Question

A dry gas occupies $125c{{m}^{3}}$ at STP. If the same gas is collected over water at ${{27}^{\circ }}C$ at a total pressure of 78 pressure of 750 Torr the volume occupied by the gas is
(Given: Aqueous tension of water at ${{27}^{\circ }}C$ is 25 Torr)

Answer 1

Hint:. -The equation used to solve the problem is $\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}$
- 1 atm = 760 Torr

Complete step by step answer:
So the question has been asked that a dry gas is occupying a volume of $125c{{m}^{3}}$ at STP, so what will the volume of the gas if the same gas is collected over water at a temperature of ${{27}^{\circ }}C$ and the total pressure is also given.
- So here to solve the equation we should know the STP conditions, i.e. is standard temperature and standard pressure conditions.
Standard temperature is taken as 273 K
Standard pressure is taken as 1 atm
So we will give these values for finding the pressure of gas collected over water.
For dry gas,
${{V}_{1}} = 125c{{m}^{3}}$, (Convert $c{{m}^{3}}$ to Litre)
${{V}_{1}} = 0.125L$
${{P}_{1}} = 1atm$, as we take the condition at STP.
${{T}_{1}} = {{0}^{\circ }}C=273K$

Now let’s find the values of the parameters of the gas collected over water.
The temperature is given, the temperature is ${{27}^{\circ }}C$. As the temperature is in degree Celsius we have to convert the value to Kelvin before solving the problems.
So, ${{T}_{2}} = {{27}^{\circ }}C = 273 + 27 = 300K$

Now the total pressure has been given as, P = Torr.
We have to find the P for dry air and let’s take it as ${{P}_{2}}$
${{P}_{2}} = 750 - 25 = 725Torr$ (as aqueous tension of water at ${{27}^{\circ }}C$ is 25 Torr)
Now we have to convert the unit of Pressure from Torr to atm, since the STP unit of Pressure is atm.
We know the relation of Torr and atm as,
1 atm = 760 Torr

Now substitute all the values in the equation,
$\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}$,
We want the value for the volume of gas collected over water, which is ${{V}_{2}}$. So write the equation for ${{V}_{2}}$.
${{V}_{2}} = \dfrac{{{P}_{1}}{{V}_{1}}{{T}_{2}}}{{{T}_{1}}{{P}_{2}}}$
${{V}_{2}} = \dfrac{1\times 0.125\times 300}{273\times 0.9539}$
${{V}_{2}} = 0.144l\,or\,144c{{m}^{3}}$

Note: The main focus and attention should be given while solving these types of the problems . Since the units are in different conditions, so before substituting the values, every value should be converted into the same units, which is the reason many get the wrong answer for these types of questions.
- And one should know the conversion units and its relation.