Question

During an adiabatic process, the temperature of 6.10 moles of a monatomic ideal gas drops from 495 degree C to 147 degree C. Find the work it does. kJ Find the heat it exchanges with the surroundings. kJ Find the change in internal energy. kJ

Answer 1

Work done adiabatic process is given as

$W=\frac{P_1V_1-P_2V_2}{\gamma -1}$

Not at 1st and 2nd state

hence

$\Delta W=\frac{nR(T_1-T_2)}{2.5 -1}=\frac{6.10\times8.31\times(495-147)}{1.5}=1.176\times10^4$ J$$

(b) Since the process is adiabatic hence heat exchange will be 0 i.e.

$\Delta Q=0$

(c)

From 1st law of thermodynam,ics

$\Delta U=\Delta Q+\Delta W=0+1.176\times10^4=1.176\times10^4$