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A non-conducting cylindrical vessel of l...

A non-conducting cylindrical vessel of length `3l` is placed horizontly & is divided into three parts by two easily moving piston having low thermal conductivity as shown in figure. These parts contains `H_(2) ,` He and `CO_(2)` gas at initial temp. `theta_(1) = 372^(@)C, theta_(2) = -15^(@)C` and `theta_(3) = 157^(@)C` respectively. If initial length and pressure of each part are l and `P_(0)` respectively. calculate final pressure and length of each part . Use : `gamma_(co_(2)) = 7//5`

Text Solution

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The correct Answer is:
`(13)/(12)P_(0) , l_(1)= 0.6l , l_(2) =1.5l , l_(3) = 0.9l`

`theta_(1) = 372^(@)C , theta_(2) = -15 ^(@)C , theta_(3)= 157 ^(@)C gamma_(CO_(2)) = (7)/(5)`
For the `H_(2)` gas `(P_(1)V_(1))/(RT_(1)) = (P_(2)V_(2))/(RT_(2)) implies (P_(0)Al)/(645R) = (P xx Al_(1))/(TR) …(i)`
For the He gas `(P_(0)Al)/(258R) = (PxxAl_(2))/(TR) .... (ii)`
For the `CO_(2)` gas `(P_(0)Al)/(430R) = (P xx Al_(2))/(TR) .... (iii)`
`implies (l_(1))/(l_(2)) = (258)/(645) = (1)/(2.5)` & `(l_(1))/(l_(3)) = (430)/(645) = (1)/(1.5)`
`implies l_(1) + l_(2) + l_(3) = 3l implies l_(1)(1 +2.5 + 1.5)=3l`
`l_(1) = 0.6 l "and" l_(2) = 2.5l_(1) = 1.5l l_(3) = 1.5 xx 0.6l = 0.9l`
For the entire system `DeltaU_(1) + DeltaU_(2) + DeltaU_(3) = 0`
`implies (n_(1)R(T-T_(1)))/(gamma_(1)-1) + (n_(2)R(T-T_(2)))/(gamma_(2)-1) + (n_(3)R(T-T_(3)))/(gamma_(3)-1) = 0`
`implies (Pxx0.6l-P_(0)l)/(((7)/(5)-1)) + ((P xx 1.5l- P_(0)l))/(((5)/(3)-1)) + (Pxx 0.9l - P_(0)l)/(((7)/(5)-1)) = 0`
`implies P =(13)/(12)P_(0)`
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