Question
Question: A volume V of a gas at a temperature \[{T_1}\] , and a pressure p is enclosed in a sphere. It is co...
A volume V of a gas at a temperature T1 , and a pressure p is enclosed in a sphere. It is connected to another sphere of volume 2V 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 T2 . The first sphere is maintained at a temperature T1 . What is the final pressure p1 within the apparatus?
Solution
If we take total number of molecules of the gas be n. of which n1 are in the larger sphere and n2 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 RT1P1V and moles in the second sphere will be 2RT2P1V . Hence, we can find the final pressure p1within the apparatus.
Complete step by step answer:
Let the total number of molecules of the gas be n. of which n1 are in the larger sphere and n2 in the smaller sphere after the stopcock is opened.
Using ideal gas equation,
PV=nRT
n=RTPV
Case I - When the stopcock is closed
Pressure enclosed in sphere =p
Temperature of sphere =T1
Volume =V
Therefore, total moles of gas
(n)=RT1PV
Case II - When the stopcock is opened.
Pressure within the apparatus, i.e., final pressure =p1
Temperature of first sphere =T1
Temperature of second sphere =T2
Volume of first sphere =V
Volume of second sphere =2V
Therefore, Moles of gas in first sphere n1=RT1p1V
and Moles of gas in second sphere
⇒n2=RT2p1(2V)
⇒n2=2RT2p1V
Now,
Total moles of gas = moles of gas in first sphere + moles of gas in second sphere
n=n1+n2
⇒RT1pV=RT1p1V+2RT2p1V
⇒p1=2T2+T12pT2
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.