Answer to Question #269907 in Molecular Physics | Thermodynamics for Hassan

Question

Air is contained in a vertical piston–cylinder assembly fitted with an electrical resistor. The atmosphere exerts a pressure of 1 bar on the top of the piston, which has a mass of 45 kg and a face area of .09 m2 . Electric current passes through the resistor, and the volume of the air slowly increases by .045 m3 while its pressure remains constant. The mass of the air is .27 kg, and its specific internal energy increases by 42 kJ/kg. The air and piston are at rest initially and finally. The piston–cylinder material is a ceramic composite and thus a good insulator. Friction between the piston and cylinder wall can be ignored, and the local acceleration of gravity is g=9.81 m/s2 . Determine the heat transfer from the resistor to the air, in kJ, for a system consisting of (a) the air alone, (b) the air and the piston

Answer 1

(a) $Q=\Delta U+p\Delta V$

$\Delta U=m_{air}\cdot\Delta _{air}u=0.27\cdot42000=11340\ (J)$

$pA=m_{piston}g+p_aA\to p=\frac{m_{piston}g+p_aA}{A}=\frac{m_{piston}g}{A}+p_a=$

$=\frac{45\cdot9.81}{0.09}+100000=104905 \ (Pa)$

$W=p\Delta V=104905\cdot0.045=4721\ (J)$

$Q=11340+4721=16061\ (J)$ . Answer

(b) $Q=W+\Delta U+\Delta PE$

$W=p_a\Delta V=100000\cdot0.045=4500\ (J)$

$\Delta U=m_{air}\cdot\Delta _{air}u=0.27\cdot42000=11340\ (J)$

$\Delta PE=m_{piston}g\Delta h=45\cdot9.81\cdot\frac{0.045}{0.09}=221\ (J)$

$Q=4500+11340+221=16061 \ (J)$ . Answer