Question

Two chambers containing \[{m_1}\]\[gm\] and \[{m_2}\]​\[gm\] of a gas at pressure \[{P_1}\] and \[{P_2}\] respectively are put in communication with each other, temperature remaining constant. The common pressure reached will be
a. \[\dfrac{{{P_1}{P_2}\left( {{m_1} + {m_2}} \right)}}{{\left( {{P_2}{m_1} + {P_1}{m_2}} \right)}}\]
b. \[\dfrac{{{P_1}{P_2}{m_1}}}{{\left( {{P_2}{m_1} + {P_1}{m_2}} \right)}}\]
c. \[\dfrac{{{m_1}{m_2}\left( {{P_1} + {P_2}} \right)}}{{\left( {{P_2}{m_1} + {P_1}{m_2}} \right)}}\]
d. \[\dfrac{{{m_1}{m_2}{P_2}}}{{\left( {{P_2}{m_1} + {P_1}{m_2}} \right)}}\]

Answer 1

Hint: According to Boyle's law, pressure is not directly proportional to volume. Instead, \[P\] and \[V\] exhibit inverse proportional relationship: Increasing the pressure results in a decrease of the volume of the gas at constant temperature. Mathematically this can be written: \[PV = cons\tan t\]
This process is also known as isothermal because temperature is constant.

Formula used:
We are using Boyle’s law for gas. According to this law, \[PV = cons\tan t\].
\[PV = nRT\]
\[ \Rightarrow PV = \left( {\dfrac{m}{M}} \right)RT\]
Where, \[\dfrac{m}{M}=\] Molecular weight of gas,
For first chamber, molecular weight= \[\dfrac{{{m_1}}}{M}\] and second chamber has molecular weight= \[\dfrac{{{m_2}}}{M}\]
\[M\] is unique for both chambers because we use the same gas in both chambers.
After combining two chambers total volume is given by addition of two volumes. Similarly total mass is given by addition of two masses. After this we can calculate common pressure by \[PV = \left( {\dfrac{m}{M}} \right)RT\] formula.

Complete step by step answer:
We know that for constant temperature, relation between pressure and volume is given by \[PV = \left( {\dfrac{m}{M}} \right)RT\]
For first chamber, \[{P_1}{V_1} = \left( {\dfrac{{{m_1}}}{M}} \right)RT\] -----(i)
Similarly, for second chamber, \[{P_2}{V_2} = \left( {\dfrac{{{m_2}}}{M}} \right)RT\] ----(ii)
When, both chambers put in contact , then total or combined volume is given by \[={V_1} + {V_2}\] and total mass is given by, \[{m_1} + {m_2}\]
Then, gas equation is given by
\[P\left( {{V_1} + {V_2}} \right) = \left( {\dfrac{{{m_1} + {m_2}}}{M}} \right)RT\] ----(iii)
From equation (i), \[{V_1} = \left( {\dfrac{{{m_1}}}{M}} \right)\dfrac{{RT}}{{{P_1}}}\]
From equation (ii), \[{V_2} = \left( {\dfrac{{{m_2}}}{M}} \right)\dfrac{{RT}}{{{P_2}}}\]
Substituting the values of $v_2$ and $v_2$ in equation (iii)
\[P\left[ {\left( {\dfrac{{{m_1}}}{M}} \right)\dfrac{{RT}}{{{P_1}}} + \left( {\dfrac{{{m_2}}}{M}} \right)\dfrac{{RT}}{{{P_2}}}} \right] = \left( {\dfrac{{{m_1} + {m_2}}}{M}} \right)RT\]
\[ \Rightarrow P\left[ {\dfrac{{{m_1}}}{{{P_1}}} + \dfrac{{{m_2}}}{{{P_2}}}} \right] = \left( {{m_1} + {m_2}} \right)\]
\[ \Rightarrow P\left[ {\dfrac{{{m_1}{P_2} + {m_2}{P_1}}}{{{P_1}{P_2}}}} \right] = \left( {{m_1} + {m_2}} \right)\]
\[\therefore P = \dfrac{{{P_1}{P_2}\left( {{m_1} + {m_2}} \right)}}{{\left( {{P_2}{m_1} + {P_1}{m_2}} \right)}}\]
Hence, common pressure is given by, \[P = \dfrac{{{P_1}{P_2}\left( {{m_1} + {m_2}} \right)}}{{\left( {{P_2}{m_1} + {P_1}{m_2}} \right)}}\].

Hence, the correct answer is option (A).

Note: Point to remember:
(1) Only one gas is filled in both chambers, so \[M\]and \[R\] are the same for both masses.
(2) On combining, the resultant volume will be the sum of two individual volumes.
(3) On combining, the resultant mass of gas is the sum of two individuals' masses.