Answer to Question #127005 in Mechanical Engineering for Zakir Hussain

Question

Two systems have the following equations of state and are contained in a closed cylinder, separated by a fixed, adiabatic and impermeable piston. N1 = 2 and N2 = 1.5 moles. The initial temperatures are T1 = 175 K and T2 = 400 K. The total volume is 0.025 m3. The piston is allowed to move and heat transfer is allowed across the piston. Determine the final temperature of the system (in Kelvin).

Answer 1

we know that, for two particular system we have,$T_1=175 K,T_2=400K$ ,$n_1=2,n_2=1.5,n=n_1+n_2$

n=3.5

$\Delta S= (\frac{1}{T_1}-\frac{1}{T_2})\Delta U$

for adiabatic process , $\Delta S=0, So,\Delta U=0 , U_1=U_2$

now we know that

$\frac{1}{T_1}=\frac{3}{2}R (\frac{n_1}{U_1})$

and also we know that fundamental equation of state for two particular system for adiabatic condition

$n_1R ln(\frac{n_1}{n})+n_2R ln(\frac{n_2}{n})=nRln(\frac{U^{1.5}V}{n^{2.5}})$

$2R ln(\frac{2}{3.5})+1.5R ln(\frac{1.5}{3.5})=3.5Rln(\frac{U^{1.5}0.025}{3.5^{2.5}})$

on solving this we get value of U as

U=11900 J

now

$\frac{1}{T}=\frac{3}{2}R (\frac{n}{U})$

$\frac{1}{T}=\frac{3}{2}8.314(\frac{3.5}{11900})$

T=` 272.6 K