Question

Given,0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10${{m}^{3}}$ at 1000 K. Given R is the gas constant in $J{{K}^{-1}}mo{{l}^{-1}}$ ​, x is:
(A) $\dfrac{2R}{4+12}$
(B) $\dfrac{2R}{4-12}$
(C) $\dfrac{4-R}{2R}$
(D) $\dfrac{4+R}{2R}$

Answer 1

Hint: As we know that ideal gas law equation is the equation of state of a hypothetical ideal gas which is the combination of empirical Charles’s law, Boyle’s law, Gay Lussac’s law and Avogadro’s law. We have to use the given values of the question in the ideal gas equation to find the answer.
Formula used:
We will use the following formula in this solution:-
PV = nRT
where,
P = pressure of the gases in the container
V = volume of the gas gases in the container
n = moles of the gases in the container
T = temperature
R = universal gas constant

Complete answer:
First all let us understand the concept of ideal gas equation followed by usage of this equation to solve the question as follows:-
-Ideal gas equation: It is the equation of state of a hypothetical ideal gas which gives the relation between variables like pressure, temperature, volume and moles or amount of gas with the combination of Charles’s law, Boyle’s law, Gay Lussac’s law and Avogadro’s law. Mathematically it can be represented as:-
PV = nRT
where,
P = pressure of the gases in the container
V = volume of the gas gases in the container
n = moles of the gases in the container
T = temperature
R = universal gas constant
-Calculation of number of moles of B:-
Values given in the question are as follows:-
P = pressure of the gases in the container = 200 Pa
V = volume of the gas gases in the container = 10${{m}^{3}}$
n = moles of the gases in the container = $(0.5+x)$ moles
T = temperature = 1000K
R = universal gas constant in $J{{K}^{-1}}mo{{l}^{-1}}$
Now put the above values in the ideal gas equation as follows:-
$\begin{align}
  & \Rightarrow PV=nRT \\
 & \Rightarrow 200Pa\times 10{{m}^{3}}=(0.5+x)mol\times R\text{ J}{{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}\times 1000K \\
 & \text{Rearrange the values:-} \\
 & \Rightarrow (0.5+x)mol=\dfrac{200Pa\times 10{{m}^{3}}}{R\text{ J}{{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}\times 1000K} \\
 & \Rightarrow (0.5+x)=\dfrac{2}{R} \\
 & \Rightarrow x=\dfrac{2}{R}-0.5 \\
 & \Rightarrow x=\dfrac{2}{R}-\dfrac{1}{2} \\
 & \Rightarrow x=\dfrac{4-R}{2R} \\
\end{align}$
-Hence x is equal to: (C) $\dfrac{4-R}{2R}$

Note:
-Always remember that units play an important role while solving the question of physical chemistry and hence try to use them along with the calculations so as to obtain an errorless result. Also kindly convert the units according to the gas constant provided, if required.