Answer to Question #213405 in Molecular Physics | Thermodynamics for Yomna

Question

 90 kJ of heat are supplied to a system at a constant volume. The system rejects 

95 kJ of heat at constant pressure and 18 kJ of work is done on it. The system 

is brought to original state by adiabatic process, i.e. (Heat add or reject of this 

process is equal zero). Determine, (i) The adiabatic work; )ii) The values of 

internal energy at all end states if initial value is 105 kJ. Draw the 

thermodynamic cycle on Pv diagram.

Answer 1

Take initial state value of internal energy =105 kJ

Given heat supplied at constant volume=90KJ

Heat rejected at constant pressure=95kJ

Work done will be=18J

Show initial value of internal

energy $U_l$ =105KJ

Process constant volume

W=0

Q=90

$90=U_m-U_l$

$U_m=U_l+90$

$U_m=105+90=195KJ$

Process constant pressure m=n

$Q=(U_m-U_n)+W_{m=n}$

$95=(U_m-U_n)-18$

$U_n-U_m=-77kJ$

$U_n=195-77=118KJ$

Adiabatic process

$Q_{n=l}=0$

$\int ∆Q=90-95=-5kJ$

$\int∆w=-18+W_{l=n}=-5$

$W_{n=l}=-5+18=13kJ$