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Two vessels A and B of equal volume (V0)...

Two vessels A and B of equal volume `(V_0)` are connected by a narrow tube which can be closed by a value. The vessels are fitted with piston which can be moved to change the volumes. Initially, the valve is open and the vessels contain an ideal gas `(C_p / C_v = gamma )` at atomspheric pressures`(P_0)` and atmoshpeheric temperature `(T_0)`. The walls of the vessels A are diathermic and those of B are adiabatic. The value is now closed and the pistons are slowly pulled out to increase the volumes of the of the vessels to dobuled the original value.(a) Find the temperatures and pressures in the two vessels. (b) The value is now opened for sufficeint time so that the gases acquire a common tempertures and pressures . Find the new values of the temperatuere and the pressures.

Text Solution

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Initial pressure = `P_0`
Initial temperature = `T_0`
Initial Volume = `V_0` `C_p/C_v = gamma`
(a) For diathermic vessel the temperature inside remains constant.
`P_1V_1 = P_2V_2`
implies `P_0V_0 = P_2 xx 2V_0`
implies `P_2 = P_0/2`,
Temperature = `T_0`
For adiabatic vessels the temperature does not remain constant. The process is adiabatic.
`T_1V_1^(gamma-1)=T_2V_2^(gamma-1)`
implies `T_0V_0^(gamma-1) = T xx (2V_0)^(gamma-1)`
implies `T_2 = T_0 xx 2^(1-gamma)`
implies `P_1V_1^gamma = P_2V_2^gamma`
`P_0V_0^gamma = P_2 xx (2V_0)^gamma`
implies `P_2 = ((P_0)/(2^gamma))`
(b) When the values are open , the temperature remains `T_0` through out
`P_1 = P_2` and `T_1=T_1=T_0`
So, `P_0 = P_1 + P_2`
`=2P_1=2P_2`
So, `P_1 = P_2 = P_0/2`
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