Respuesta :
Solution:
The data provided in the question are :
[tex]$V_1 = V_2 = 43.8\ L$[/tex]
= [tex]$ \frac{43.8}{1000}\ m^3$[/tex]
[tex]$ P_1 = 1.1\ MPa$[/tex] and [tex]$ P_2 = 0$[/tex]
Initial pressure of neon = 1.1 MPa
Final Pressure = 346 kPa
Initial temperature of neon = 298 K
[tex]$P_1V_1=mRT_1$[/tex]
[tex]$ 1.1 \times 10^6 \times \frac{43.8}{1000} = m \times \frac{8314}{MM}\times 298$[/tex]
Molecular mass of neon = 20.1797 g/mole
m = 0.3924 kg
For final temperature:
[tex]$P_fV_f=mRT_f$[/tex]
[tex]$V_f = 2 \times \frac{43.8}{1000}$[/tex]
[tex]$ 346 \times 1000 \times 2 \times \frac{43.8}{1000} = m \times \frac{8314}{20.1797} \times T_f$[/tex]
[tex]$ \therefore T_f = 187.48\ K$[/tex]
a). From first law of thermodynamics :
δQ = δU + δW
Tds = dU + PdV
or dH = dTs + VdP
As system is insulator, Tds = 0
[tex]$ \Delta H = \left( \frac{P_1V_1-P_2V_2}{\gamma - 1} \right)^{\gamma}$[/tex] as [tex]$PV^{\gamma}$[/tex] = constant
[tex]$P_1V_1^{\gamma}= PV^{\gamma}$[/tex]
[tex]$V= \left( \frac{P_1V_1^{\gamma}}{P} \right)^{\frac{1}{\gamma}}$[/tex]
Substituting in VdP and integrating, the above equation is obtained.
So, γ = 1.67 (mono atomic neon)
[tex]$ \Delta H = 1.67 \times \frac{(1.1 \times 10^6 \times 0.0438 - 346 \times 10^3 \times 2 \times 0.0438)}{1.67-1}$[/tex]
[tex]$ \Delta H = 44942.63\ J$[/tex]
[tex]$ \Delta H = 44.942\ kJ$[/tex]
b). Easier way is :
[tex]$ \Delta H = mC_P\Delta T$[/tex]
[tex]$ \Delta H = 0.3924 \times C_P(T_f-T_1)$[/tex]
[tex]$C_P = \frac{\gamma R}{\gamma-1}$[/tex]
[tex]$= \frac{1.67 \times 8314}{0.67 \times 20.1797}$[/tex]
= 1026.92 J/kg-K
[tex]$ \Delta H = 0.3924 \times 1026.92 (187.48-298)$[/tex]
[tex]$ = -44.585\ kJ$[/tex]
The negative sign indicates decrease in enthalpy.
The answer by easier way is very near to the value in part (a).
Error (%) = [tex]$ \frac{44.942-44.585}{44.942} \times 100$[/tex]
= 0.015 % (which is negligible)
Therefore, both the answers are same.
