a sample of gas is heated by three different methods from same initial state as shown. in each methods heat supplied is the same. in(1) piston moves up by some amount. in (2) piston moves down and in (3) piston doesnt move. specific heats of the gas calculated in each of the methods to be C1, C2, C3 we need to find which specific heat is greatest and which lowest.. pic is attached wherein figure and options are given

Ayush_Deep
120 Points
3 years ago
In first case work done W = positive
In second case work done is negative
In third case it is zero .
Q= U +W
As Q is constant ,
In first case w is positive so U = Q-w
And change in temperature in 1 will be less than in 3(w=0)
In second case change in temperature will be greater than third case.
As we know, for same heat , heat capacity in inversely proportional to change in temp
So  heat capacity of 1 will be more than 3
And heat capacity in 2 will be less than 3
Vikas TU
14149 Points
3 years ago
Dear student
As the gas heats up, its pressure should rise, which in turn displaces the piston in the outward direction. This is positive work, i.e. work being done by the system. Since the piston moves out, the gas mixture does work on ts surroundings.
Good Luck