Question

One mole of $ {O_2} $ ​ gas is contained in a box of volume $ V = 2{m^3} $ at pressure $ {P_ \circ } $ ​ and temperature $ 300K $ . The gas is now heated to $ 600K $ and the molecules now get dissociated into oxygen atoms. The new pressure of the gas is
 $ \left( A \right){\text{ }}{{\text{P}}_ \circ } $
 $ \left( B \right){\text{ 2}}{{\text{P}}_ \circ } $
 $ \left( C \right){\text{ 4}}{{\text{P}}_ \circ } $
 $ \left( D \right){\text{ 8}}{{\text{P}}_ \circ } $

Answer 1

Hint: Here, in this question, we have to find the new pressure of the gas and for this, we will use the concept of the gas equation, and it is given by $ \dfrac{{{P_2}{V_2}}}{{{T_2}}} = \dfrac{{{P_1}{V_1}}}{{{T_1}}} $ . From this equation, we will get the ratio of pressure and by substituting the values we will get the result.

Formula used:
The gas equation will be given by,
 $ \dfrac{{{P_2}{V_2}}}{{{T_2}}} = \dfrac{{{P_1}{V_1}}}{{{T_1}}} $
Here,
 $ {P_1} $ , will be the initial pressure,
 $ {P_2} $ , will be the final pressure,
 $ {V_1} $ , will be the initial volume,
 $ {V_2} $ , will be the final volume,
 $ {T_1} $ , will be the initial temperature,
 $ {T_2} $ , will be the final temperature.

Complete step by step answer
Here in this question, we have the values given as
 $ {T_2} = 600K $
 $ {T_1} = 300K $
And as we know that when the molecules break into atoms, the number of moles will be equal to half of it. So, from the gas equation, we know that
 $ \Rightarrow pV = nRT $
And here in this question, the volume is constant. Here in this question, we assume initial pressure $ {P_1} $ will be equal to $ {P_ \circ } $ .
Hence, the above equation can be written as
 $ \Rightarrow \dfrac{{{P_2}{V_2}}}{{{T_2}}} = \dfrac{{{P_1}{V_1}}}{{{T_1}}} $
And also it can be written as
 $ \Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{{V_1}}}{{{V_2}}} \cdot \dfrac{{{T_2}}}{{{T_1}}} $
And also the volume will become half.
So, we get
 $ \Rightarrow {V_2} = \dfrac{1}{2}{V_1} $
Now substituting the values in the above equation we will get the equation as
 $ \Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{{V_1}}}{{\dfrac{1}{2}{V_1}}} \cdot \dfrac{{{T_2}}}{{{T_1}}} $
And on solving the above equation we get
 $ \Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = 2 \cdot \dfrac{{{T_2}}}{{{T_1}}} $
Now substituting the known values, we will get the equation as
 $ \Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = 2 \cdot \dfrac{{600K}}{{300K}} $
And on solving the division and the multiplication, we will get the equation as
 $ \Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = 4 $
Hence, the pressure will be equal to four times of the old pressure.
Therefore, the new pressure will be equal to $ 4{P_ \circ } $ .
Hence, the option $ \left( C \right) $ is correct.

Note
This question can also be solved by using the other concept. As we know that when the gas will break the moles will be two times the initial,
So from the gas equation, it can be written as $ p \prec nT $ And from this we will have the equation $ \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{2{n_1}}}{{{n_1}}} \cdot \dfrac{{{T_2}}}{{{T_1}}} $ and on substituting the values and solving it we will get the exact same answer. So in this way also we can solve it.