Answer to Question #299029 in Molecular Physics | Thermodynamics for Romer Zamora

Question

one kg of air expand at a constant temperature from a pressure of 820 kpa and a volume of 3 m³ to a pressure of 220 kpa. determine.

A. Change of Internal Energy

B. Change of Enthalpy

Answer 1

$W=P∆V$

$W=P(V_2-V_1)$

$\frac{P_1}{P_2}=\frac{V_1}{V_2}\\V_2=\frac{P_2}{P_1}V_1$

$V_2=\frac{820}{220}\times3=11.18m^3$

$$W=P(V_2-V_1)=820\times10^3\times(11.18-3)=6.7076KJ$$

∆Q=mRT

$∆Q=1\times8.31\times273=2.278KJ$

Thermodynamic first law

$∆U=∆Q-∆W$

$$∆U=(2.278-6.7076)KJ=-4.4296KJ$$

Enthalpy change

$∆H=U+P∆V$

$$∆H=-4.4296+6.7076=2.278KJ$$