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One mole of a certain ideal gas is conta...

One mole of a certain ideal gas is contained under a weight-less piston of a vertical cylinder at a temperature `T`. The space over the piston opens into the atmosphere. What work has to be performed on order to increase isothermally the gas volume under the piston `n` times by slowly raising the piston ? The friction of the piston against the cylinder walls is negligibly small.

Text Solution

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As gas expands from its initial volume (say V) to `eta` time of initial volume (`etaV`) we can say that work is done by the gas in this expansion.As this expansion is isothermal,workdone by the gas is simply given as
WnRT ln `(V_2/V_1)`
Here n=1 mole , `V_2=eta V_1` thus
`W=RT ln eta` ...(3.52)
As gas is expanding,it is doing work and some external agent is pulling the piston up to increase the volume of gas we can say that gas is supporting the external agent and atmosphere is opposing this expansion thus work is done on atmosphere. As atmospheric pressure to be constant, and change in volume of gas is from V to `etaV`, the work done on atmosphere is
`W_"atm=-P(DeltaV)`
`=PV(1-eta)`
Initially the piston was in equilibrium thus gas pressure was equal to atmospheric pressure so, we have
PV=RT [As n=1 mole]
or `W_"atm"=-RT(eta-1)` ...(3.53)
If `W_"ext"` is the work done by external agent in pulling the gas here
we must have
`|W_"ext"|+|W_"gas"|=|W_"atm"|` ...(3.54)
`W_"ext"=W_"atm"-W_"gas"`
`W_"ext"=RT(eta-1)-RT ln eta`
Here equation-(3.54) the basic equation we've encountered several times in mechanics as if somework is being done then at least two objects are to be involved in the system, one who is doing the work (energy supplier) and the other on which work is done (energy accepter). Only one body can never do anywork. Thiswe'll discuss in next section and then we'll take some more examples on such concepts.
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