Question

Twenty moles of a monatomic ideal gas (? = 5/3) undergo an adiabatic process. The initial pressure is 400 kPa and the initial temperature is 450 K. The final temperature of the gas is 320 K. In the situation above, the change in the internal energy of the gas, in kJ, is closest to:

Answer 1

$W=\frac {P_fV_f-P_iV_i}{1-\gamma}$

$W=\frac {nRT_f-nRT_i}{1-\gamma}$

$W=nR\left (\frac {T_f-T_i}{1-\gamma} \right )$

$W=20\times 8.314\left (\frac {320-450}{1-\frac {5}{3}} \right )$

$\Delta Q =W+\Delta U$ (1st law of thermodynamics)

$\Delta Q =0$ (Adiabatic process)

$\Delta U=-W=-32.42\: kJ$