JEE Advanced 2024 Heat & Thermodynamics Nightmare: Tackle it Like a Pro

[@r4music220]

An ideal mono-atomic gas is confined in a cylinder with the help of a piston that can slide inside the cylinder without friction. The cylinder is placed on a horizontal platform in a vacuum chamber. The walls of the cylinder and the piston are made of heat insulating materials. Initially the temperature of the gas is T and the piston stays in equilibrium. The temperature of the gas is quickly increased n times and then the gas is allowed to expand. If the piston rises and again acquires an equilibrium state, find temperature of the gas in this new equilibrium state. Please deal with problems like these also, which are not of the regular type. 🙏

[@basics1729]

Thats a great suggestion 👍. I am sure gonna take up these kind of questions too in the upcoming videos.

[@sukhjotsingh799]

Broo... This is so so op question... But it's solution is a bit trash... Like they don't consider energy change when n temperature changes... It was in sri chaitanya test..

[@basics1729]

I am gonna try to cover this in the best possible manner 👍

[@r4music220]

@@basics1729 thank you bro😀

[@r4music220]

@@sukhjotsingh799 I got logic while attempting test that it is irreversible idk concepts though

[@pranavdewangan8644]

SIr , I understood the complete solution but things which i want to clear :1) why don't we take work done by gases in the 1st law (work done by system),2)you told we have to write work by system - the system contains spring and gases , so work by spring(Ui -Uf)(work by conservative) - should be negative (so is spring the part of system) . I think i am not able to get the concept of work done by system

[@basics1729]

I am answering as per your points - 1. In this particular case, we are actually taking work done by gases in the 1st law but suppose a system having no gas within it and some other fluid or maybe some solid. That is why its better to write work done by the system instead of work done by the gas. 2. Yes, here the system consists of both the springs and the gas and that piston too. But the phenomenon that is triggering the springs to compress or expand is basically heat transferred to the gas which is getting expanded and due to which springs are doing work. So, either you take the work done by the gas or -ve of work done by the spring but not both as you will get zero by adding both because both are one and the same thing (action-reaction) kind of thing. I hope you get my point. You can even see other comments below based on the same doubt.

[@sukhjotsingh799]

Sir... Work by spring to negative aa rha h na? Aapne +ve 1/2kx^2 kyu likha?

[@basics1729]

Actually system is doing work when Q heat is given to it from the right side and as a result the potential energy is getting stored in the spring.

[@sukhjotsingh799]

@@basics1729 can we say that total work was zero.. Work by gases + wspring = 0 ...?

[@basics1729]

Yes ofcourse, this is because when Q amount of heat is transferred to the gas then as a result of pressure on the piston, spring will get expanded on the right hand side of the partition and compressed on the left hand side and so gas is doing work upon the system. But springs in turn will try to regain its original length and thats why the force acts in the opposite direction to their displacements. That is why the work done by the gas and springs are opposite and getting cancelled when you add them up.

[@basics1729]

See, because gas is doing work upon the system thats why springs are getting disturbed but spring will also try to regain its natural length and you are considering this spring force only while getting work done by the springs and that is why you are getting zero as work done. Don't consider like this, you have to be very clear about the system consideration otherwise you are gonna commit mistake like this by considering the same action-reaction kind of work. I hope you get it 👍