A fixed mass of an ideal gas is compressed slowly so that its volume halves and its pressure doubles.
Which of the following statements about the gas after this change is/are correct?
1 The final temperature of the gas is the same as before the change.
2 The final internal energy of the gas is the same as before the change.
3 The final mean kinetic energy of the particles of the gas is the same as before the change.
1, 2 and 3
2 and 3 only
1 and 3 only
none of them
Hi so the answer to the question is A. So we know that pressure was double as volume was halfed, therefore P1V1=P2V2 and therefore there could not bee any increase in temprature, and the only thing that increases kinetic energy for gases is an increase in temprature. Therefore, statement 1 and 3 are both correct.
The part I am stuck on is statement 2. As there is mechanical work done on the system and as delta U = q + w, then the internal energy of the system must have changed? or is there no mechanical work done on the system by compression it?
alright so i found this online which explains why work was not done on the system:
To calculate how much work a gas has done (or has done to it) against a constant external pressure, we use a variation on the previous equation:
work = - P(pressure) * change in (volume)
as the pressure was doubled and volume was halfed there was no net work done on the system or by the system.
LATM is the unit for work and 1 LATM = 101.3J.
The source is khan academy:
all of these are related with temperature. think very straight if there is no change in temperature then there is no change in temperature. internal energy literally is a way of saying temperature and kinetic energy means temperature as well. no need for flashy formulas etc. kinetic e internal energy all mean the same thing.
pV=nRT surely just means that if P doubles and V halves that the nRT remains the same right? With T being constant - thats what I gathered I think?
yes exactly. no changes in T explains all
in the videos I had watch on internal energy Delta U = q + w. The q is change in temperature but the w is more based on mechanical work which is why I was confused.