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At a given temperature, the pressure in ...

At a given temperature, the pressure in a oxygen cylinder is 10.3 atm. At the same temperature the pressure in another oxygen cylinder of volume 1/3rd of the first cylinder is 1.1 atm. Keeping temperature constant, if the two cylinders are connected, then what will be the pressure of `O_(2)` gas in the system?

Text Solution

Verified by Experts

Suppose the volumes of the first cylinder=VL.
So, the volume of another cylinder `(V)/(3)L`.
In case of first cylinder: if the number of moles of `O_(2)` gas be `n_(1)`, then according to the equation PV=nRT
`10.3atmxxVL=n_(1)xx0.0821L*atm*mol^(-1)*K^(-1)xxTK`
`therefore n_(1)=(125.45xxV)/(T)` mol
In case of second cylinder: if the number of moles of `O_(2)` gas be `n_(2)`, then according to the equation PV=nRT
1.1 atm `xx(V)/(3)L=n_(2)xx0.0821L*atm*mol^(-1)*K^(-1)xxTK`
`therefore n_(2)=(4.46)/(T)xx`V mol
If two cylinders are connected together, then the total volume of the system `=(V+(V)/(3))L=(4)/(3)` VL and the total number of moles of `O_(2)` gas `=n_(1)+n_(2)`.
Applying equation PV=nRT, we obtain
`Pxx(4)/(3)VL=(n_(1)+n_(2))RT=((125.45xxV)/(T)+(4.46xxV)/(T))xx0.0821xxT`
or, `P=(3)/(4)xx((125.45+4.46))/(T)xx0.0821xxT`
`=(3)/(4)xx129.61xx0.0821atm=7.99` atm
`therefore` pressure of the `O_(2)` gas in the system will be 7.99 atm.
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