Question: One litre of oxygen at a pressure 1atm and two litre nitrogen at a pressure of 0.5atm, are induced i...

Question

One litre of oxygen at a pressure 1atm and two litre nitrogen at a pressure of 0.5atm, are induced into a vessel of volume 1L. If there is no change in the temperature, the final pressure of the mixture of gas (in atm) is:
A. 1.5
B. 2.5
C. 2
D. 4

Answer 1

Solution

In the given question, you can use the ideal gas equation. Also, the temperature is said to remain constant. This provides two constant entities namely temperature and real gas constant. This simplifies our calculations as we won’t have to assume any new variable or values for this question.

Formula used:
In order to solve the given question, we will be using the ideal gas equation, i.e.,
$pV=nRT$

Complete answer:
To solve the given question, let's start from the ideal gas equation,
$pV=nRT$
Where, p is pressure of the ideal gas
V is the volume of the ideal gas or the container in which it is being stored
n is amount of substance
R is ideal gas constant, $R=8.3145J.mo{{l}^{-1}}.K_{{}}^{-1}$
And T is the temperature of the gas
Now, for oxygen,
$p=1atm$
$V=1L$
And let $n={{n}_{{{O}_{2}}}}$
By using the above given parameters in Ideal gas equation,
$\begin{aligned} & 1\times 1={{n}_{{{O}_{2}}}}RT \\\ & \\\ & \therefore {{n}_{{{O}_{2}}}}=\dfrac{1}{RT} \\\ \end{aligned}$

Now for the nitrogen gas,
$p=0.5atm$
$V=2L$
And let $n={{n}_{{{N}_{2}}}}$

Again, using the above parameters in Ideal Gas equation

& 0.5\times 2={{n}_{{{N}_{2}}}}RT \\\ & \\\ & \therefore {{n}_{{{N}_{2}}}}=\dfrac{1}{RT} \\\ \end{aligned}$$ Now, if they are mixed, Let the new pressure be ${{p}_{new}}$ Now, By the ideal gas equation, ${{p}_{new}}{{V}_{new}}={{n}_{total}}RT$ Since, it is given to us that ${{V}_{new}}=1L$ and temperature remain constant ${{p}_{new}}\times 1={{n}_{total}}RT$ $\Rightarrow {{p}_{new}}=\left( \dfrac{1}{RT}+\dfrac{1}{RT} \right)RT$ $\Rightarrow {{p}_{new}}=\left( \dfrac{2}{RT} \right)RT$ $\Rightarrow {{p}_{new}}=2atm$ So, the new pressure of the mixture of the gases will be 2atm. **So, the correct answer is “Option C”.** **Note:** Sometimes, the gases given in the question might not be the ideal gases. In such cases the Ideal Gas equation can not be used. You will have to use Real Gas Equation for such question or Van Der Waals equation, i.e., $$~(P~+~\dfrac{a{{n}^{2}}}{{{V}^{2}}})(V-nb)=~nRT$$