30Swordfish.. Nowhere is it mentioned that it is a reversible process. So the formula for work done is redundant.
For irreversible process,
W=-PextΔV
so, ΔE= q - PextΔV
Calculate the increase in internal energy of 1 kg of water at 100 degrees Celsius when it is converted into steam at the same temperature and at 1atm. The density of water and steam are 1000kg/m3 and 0.6kg/m3 resp. The latent heat of vaporisation of water is 2.25 x 106 J/kg
Attempt-
The heat given to convert 1 kg of water to steam is 2.25 x 106 J.
I don't understand whether work is done BY the system or ON the system in increasing its volume.
The work done (by or on) the system is given by
W=nRTln(V2/V1) as the temperature remains same.
On adding both the values I got a wrong answer.
The correct answer is 2.08 x 106 J.
Please explain.
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5 Answers
30
1It is given that 'at the same temperature' (in the question).
So conclusion is that it is an isothermal process.
Why can't we apply W=nRTln(V2/V1)?
30See, there's no doubt that it is an isothermal process. But, if you look carefully at the question, it is implied that is an irreversible process. Because see, they have mentioned that the process takes place at constant pressure of 1atm. In a reversible process, at every step internal and external pressure are at equilibrium. So, the given process cannot be reversible. And the formula for work done, W=nRTln(V2/V1) can be applied only for reversible process.
Thats why the difference in answer!
1Thanks alot,
One thing- Evaporation of water is a reversible process. Moreover, evaporation takes place only when vapour pressure of water equalises the atmospheric pressure (or external pressure which is 1atm in this case). So the pressure of the system and surroundings is also in equilibrium.
PLease explain I am desperate.
