Question

Express the change in
internal energy of a system when (i) No heat is absorbed

by the system from
the surroundings, but work (w) is done on the system. What

type of wall does the
system have?(ii) No work is done on the system, but q

amount of heat is
taken out from the system and given to the surroundings.

What type of wall
does the system have? (iii) w amount of work is done by the

system and q amount
of heat is Supplied to the system. What type of system

would it be?

Answer 1

First law of thermodynamics: ΔU = Q - W
where ΔU = change in Internal energy
            Q = Heat added to the system
            W = work done by the system

i)Q= 0 ⇒ΔU = 0 - (-W)
     ⇒ ΔU = W, Wall is adiabatic.
ii)W= 0 ⇒ ΔU = (-Q) -0
⇒ΔU = -Q; Wall is non-adiabatic or thermally conducting
iii)ΔU = Q - W

Answer 2

Answer:

(i) adiabatic wall, (i)  conducting wall and (iii) closed system

Explanation:

According to First law of thermodynamics, change in internal energy is equal to heat energy added to the system minus work done by the system, i.e., $\Delta U = Q - W$

(i) Given no heat is absorbed by the system, $Q= 0$

  and work is done on the system,  $W =-W$

The change in internal energy of a system is

                 $\Delta U = 0 - (-W)$

         $\implies \Delta U = W$

As no heat is allowed from the surroundings, the wall is adiabatic.

ii) Given no work is done on the system, $W =0$

             but heat is given to the surroundings, $Q= -Q$

 The change in internal energy of a system is

                 $\Delta U = -Q -0$

         $\implies \Delta U = -Q$

As heat is allowed,  the wall is thermally conducting.

iii) Given work is done by the system, $W =+W$

             but heat is given to the system, $Q=+Q$

 The change in internal energy of a system is

                 $\Delta U = Q +W$

   So, the system is a closed system.