Question

Two containers of equal volume contain the same gas at pressure \[{{p}_{1}}\]and \[{{p}_{2}}\]and absolute temperature \[{{T}_{1}}\] and \[{{T}_{2}}\], respectively. On joining the vessels, the gas attains a common pressure \[P\] and common temperature \[T\]. The ratio \[{P}/{T}\;\] is equal to
A.\[\dfrac{{{p}_{1}}}{{{T}_{1}}}+\dfrac{{{p}_{2}}}{{{T}_{2}}}\]
B.\[\dfrac{1}{2}\left( \dfrac{{{p}_{1}}}{{{T}_{1}}}+\dfrac{{{p}_{2}}}{{{T}_{2}}} \right)\]
C.\[\dfrac{{{p}_{1}}{{T}_{2}}+{{p}_{2}}{{T}_{1}}}{{{T}_{1}}+{{T}_{2}}}\]
D.\[\dfrac{{{p}_{1}}+{{p}_{2}}}{{{T}_{1}}+{{T}_{2}}}\]

Answer 1

Hint:Let's make use of the formula to find out the number of molecules in a gas and try to solve it. N number of molecules in a gas of volume V is given by
\[N=\dfrac{PV}{kT}\] where P is the pressure of the gas, T is the temperature of the gas and k is a constant, called Boltzmann constant.


Formula used:
The ideal gas law:
\[PV=NkT\]
P is the pressure of the gas.
V volume of gas.
N is the total number of molecules in the gas.
k is the Boltzmann’s constant, and
T is the absolute temperature of the gas.


Complete step-by-step solution:
Let the volume of gas in each container be V.
Assuming the first container contains \[{{N}_{1}}\] number of molecules of the gas, at pressure \[{{p}_{1}}\] and absolute temperature \[{{T}_{1}}\].
Then, from ideal gas equation:
\[\begin{gathered}
  & {{p}_{1}}V={{N}_{1}}k{{T}_{1}} \\
 & \dfrac{{{p}_{1}}V}{k{{T}_{1}}}={{N}_{1}} \\
\end{gathered}\]
Similarly, assuming the second container contains \[{{N}_{2}}\] number of molecules of the gas, at pressure \[{{p}_{2}}\] and absolute temperature \[{{T}_{2}}\].
Then, from ideal gas equation:
\[\begin{gathered}
  & {{p}_{2}}V={{N}_{2}}k{{T}_{2}} \\
 & \dfrac{{{p}_{2}}V}{k{{T}_{2}}}={{N}_{2}} \\
\end{gathered}\]
Now, on joining the two vessels, the the gas attains a common pressure \[P\] and common temperature \[T\].
The total volume of the gas \[=V+V=2V\]
The total number of molecules in the gas \[={{N}_{1}}+{{N}_{2}}=N(\text{say)}\] [the sum of number of molecules in the two containers]
From ideal equation, the total number of molecules in the gas of volume \[\text{2}V\]at pressure \[P\] and temperature \[T\] is given by:
\[\begin{gathered}
  & P(2V)=NkT \\
 & \dfrac{PV}{kT}=N \\
\end{gathered}\]
Now \[N={{N}_{1}}+{{N}_{2}}\]
Therefore,
\[\begin{gathered}
  & \dfrac{P(2V)}{kT}=\dfrac{{{p}_{1}}V}{k{{T}_{1}}}+\dfrac{{{p}_{2}}V}{k{{T}_{2}}} \\
 & \dfrac{V}{k}\left( \dfrac{2P}{T} \right)=\dfrac{V}{k}\left( \dfrac{{{p}_{1}}}{{{T}_{1}}}+\dfrac{{{p}_{2}}}{{{T}_{2}}} \right) \\
 & \dfrac{2P}{T}=\dfrac{{{p}_{1}}}{{{T}_{1}}}+\dfrac{{{p}_{2}}}{{{T}_{2}}} \\
 & \dfrac{P}{T}=\dfrac{1}{2}\left( \dfrac{{{p}_{1}}}{{{T}_{1}}}+\dfrac{{{p}_{2}}}{{{T}_{2}}} \right) \\
\end{gathered}\]


Thus, option B is the correct answer.
Additional information: Another form of the ideal gas law is \[PV=nRT\] and is obtained by writing \[N=n{{N}_{A}}\] in \[PV=NkT\] where \[{{N}_{A}}\] is the Avogadro number and n is the number of moles of the gas. \[R={{N}_{A}}k\] and is called the universal gas constant.


Note: The given problem can also be solved taking into consideration the number of moles of the gas in the two containers using the equation \[PV=nRT\] (in the same way as solved above).