Question

The size of container B is double that of A and gas in B is at double the temperature and pressure than that in A. Find the ratio of molecules in the two containers.

Answer 1

Hint
According to the question there are two containers . we need to find the ratio of two types of molecule,we use the ideal gas equation.

Complete answer:
The number of molecules= A × n..........[A=avogadro number, n=number of moles]
According to the question the volume , pressure, temperature of container B is twice of container A. So,
$\begin{gathered}
  {V_B} = 2{V_A} \\
  {T_B} = 2{T_A} \\
  {P_B} = 2{P_A} \\
\end{gathered} $
Now, according to the ideal gas equation,
$PV=nRT$
For,container A
$\begin{gathered}
  {P_A}{V_A} = {n_A}R{T_A} \\
  \therefore {n_A} = \dfrac{{{P_A}{V_A}}}{{R{T_A}}} \\
\end{gathered} $
For, container B,
$\begin{gathered}
  {P_B}{V_B} = {n_B}R{T_B} \\
  \therefore {n_B} = \dfrac{{{P_B}{V_B}}}{{R{T_B}}} \\
\end{gathered} $
Now we can put the values of $P_B$,$T_B$,$V_B$
$\begin{gathered}
  \therefore {n_B} = \dfrac{{4{P_A}{V_A}}}{{2R{T_A}}} \\
  \therefore {n_B} = \dfrac{{2{P_A}{V_A}}}{{R{T_A}}} \\
\end{gathered} $
So, the the ratio of total number of molecule is,
$\dfrac{{{n_B}}}{{{n_A}}} = \dfrac{{\dfrac{{2{P_A}{V_A}}}{{R{T_A}}}}}{{\dfrac{{{P_A}{V_A}}}{{R{T_A}}}}} = \dfrac{2}{1}$
So the ratio of the molecules is = 2:1.

Note
The term ideal gas refers to a hypothetical gas composed of molecules which follow a few rules: Ideal gas molecules do not attract or repel each other. The only interaction between ideal gas molecules would be an elastic collision upon impact with each other or an elastic collision with the walls of the container.