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A piston can freely move inside a horizo...

A piston can freely move inside a horizontal cylinder closed from both ends. Initially, the piston separates the inside space of the cylinder into two equal parts each of volmek `V_0` in which an ideal gas is contained under the same pressure `p_0` and at the same temperature. What work has to be performed in order to increase isothermally the volume of one part of gas `eta` times compared to that of the other by slowly moving the piston ?

Text Solution

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As the piston is displaced externally some external work is done in the process. If piston is displaced toward right the gas on left side expands and does some work. Similarly gas on right is compressed and on it work is done.As discussed earlier, here we have
`|W_"gas in left part|+|W_"ext"|=|W_"gas i n right part"|`...(3.61)
Here `|W_"gas on left"|`= magnitude of work done by gas in left part
`|W_"ext"|`= magnitude of work done by external agent
`|W_"gas in right part"|`=magnitude of work done on gas in right part
It is given that initial volume of both the parts is `V_0` and in the process final volume of one part is `eta` times that of the other part.If the final volume of right part is V then that of left part will become `etaV`. As total volume of container is `2V_0`, then we have
`V+etaV=2V_0`
or `V=(2V_0)/(eta+1)`
For gas in left part work done by gas in isothermal expansion is
`W_"by gas"=nRT ln V_2/V_1`
`rArr =P_0V_0 ln (etaV)/V_0` [As from gas law for gas on left part `P_0V_0=nRT` ]
`rArr =P_0V_0 ln((2eta)/(eta_1))`...(3.62)
Similarly for gas in right part,workdone on the gas in isothermal compression is
`W_"on gas"=nRT ln V_2/V_1`
`rArr =P_0V_0 ln V/V_0`
`rArr =P_0V_0 ln (2/(eta+1))`
`rArr =-P_0V_0 ln ((eta+1)/2)`
Now from equation-(3.61) , (3.62) and (3.63), we have
`|W_"ext"|=|W_"by gas in left part"|+|W_"on gas in right part"|`
`=-P_0V_0 ln(2eta)/(eta+1) +P_0V_0 ln((eta+1)/2)`
`=P_0V_0 ln[(eta+1)^2/(4eta)]`
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