Question

A volume V of a gas at a temperature \[{T_1}\] ​, and a pressure p is enclosed in a sphere. It is connected to another sphere of volume $\dfrac{V}{2}$ by a tube and stopcock. The second sphere is initially evacuated and the stopcock is closed. If the stopcock is opened the temperature of the gas in the second sphere becomes \[{T_2}\] . The first sphere is maintained at a temperature \[{T_1}\] ​. What is the final pressure \[{p_1}\] ​within the apparatus?

Answer 1

Hint: If we take total number of molecules of the gas be n. of which \[{n_1}\] are in the larger sphere and \[{n_2}\] in the smaller sphere after the stopcock is opened and Using ideal gas equation, \[PV = nRT\] . Total moles of gas (n) in first sphere will be $\dfrac{{{P_1}V}}{{R{T_1}}}$ and moles in the second sphere will be $\dfrac{{{P_1}V}}{{2R{T_2}}}$ . Hence, we can find the final pressure \[{p_1}\]within the apparatus.

Complete step by step answer:
Let the total number of molecules of the gas be n. of which \[{n_1}\] are in the larger sphere and \[{n_2}\] in the smaller sphere after the stopcock is opened.
Using ideal gas equation,
\[PV = nRT\]
$n = \dfrac{{PV}}{{RT}}$
Case I - When the stopcock is closed
Pressure enclosed in sphere \[ = p\]
Temperature of sphere \[ = {T_1}\]
Volume \[ = V\]
Therefore, total moles of gas
$(n) = \dfrac{{PV}}{{R{T_1}}}$
Case II - When the stopcock is opened.
Pressure within the apparatus, i.e., final pressure \[ = {p_1}\]
Temperature of first sphere \[ = {T_1}\]
Temperature of second sphere \[ = {T_2}\]
Volume of first sphere \[ = V\]
Volume of second sphere \[ = \dfrac{V}{2}\]
Therefore, Moles of gas in first sphere \[{n_1} = \dfrac{{{p_1}V}}{{R{T_1}}}\]
and Moles of gas in second sphere
\[ \Rightarrow {n_2} = \dfrac{{{p_1}\left( {\dfrac{V}{2}} \right)}}{{R{T_2}}}\]
\[ \Rightarrow {n_2} = \dfrac{{{p_1}V}}{{2R{T_2}}}\]
Now,
Total moles of gas = moles of gas in first sphere + moles of gas in second sphere
\[n = {n_1} + {n_2}\]
$ \Rightarrow \dfrac{{pV}}{{R{T_1}}} = \dfrac{{{p_1}V}}{{R{T_1}}} + \dfrac{{{p_1}V}}{{2R{T_2}}}$
$ \Rightarrow {p_1} = \dfrac{{2p{T_2}}}{{2{T_2} + {T_1}}}$

Note: We can define an ideal gas as a hypothetical gaseous substance whose behaviour is independent of attractive and repulsive forces and can be completely described by the ideal gas law.