Question

in changing the state of gas adiabatically from any ko William state a to another equilibrium state be an amount of work equal to 22.3 Jaaye is done on the system if the gas is taken from state a to b y a process in which the net heat absorbed by the system is 9.35 cal how much is the net work done by the system in the later case ​

Answer 1

given work done = -22.3J

work done is taken negative as work is done on the system

in an adiabatic system delQ= 0

using first law of thermodynamics,

U=Q-W

U= 22.3J

for another process between state A and state B,

heat absorbed = 39.18 J

change in internal energy between two state by a different parts are equal.

from first law of thermodynamics,

U=Q-W

U=16.9J

Answer 2

Explanation:

Given, the work done on the system is

22.3 and heat absorbed by the system is

and 9.35 cal.

This is an adiabatic process. so change is heat is 0(∆Q = 0).

∆W = -22.3 J(Negative sign indicates that work is done on the system)

So, from the first law of thermodynamics,

∆Q = ∆U + ∆W

=> ∆U = 22.3 J

Now,

When the system goes from state A to state by absorbing of heat, then

∆Q = 9.35 cal

= 9.35 * 4.2

= 39. 3 J

Then,

Heat absorbed ∆W = ∆Q - ∆U

=> 39.3 - 22.3

=> 17 J

The network done by the system in the latter cas is 17.0 J.

Hope it helps!