Answer to Question #273790 in Molecular Physics | Thermodynamics for KIM

Question

A perfect gas has a value of R = 316 J/kg-K and k = 1.26. If 21 kJ of heat are added to 2.5 kg of this gas during an isometric process when the initial temperature is 32 deg. C, determine:

(a) the final temperature in deg. C

(b) the change in enthalpy in kJ

(c) the change in entropy in kJ/K

Answer 1

Solution;

Given;

R=316J/kgK

k=1.26

Q=21kJ

m=2.5kg

$T_1=32°c=305K$

(a)

To find the final temperature;

$Q=mc_p(T_2-T_1)$

$c_p=\frac{kR}{k-1}=\frac{1.26×316}{0.26}=1531.38J/kgK$

Therefore;

$21000=2.5×1531.38(T_2-305)$

$T_2=5.485+305=310.485K$

$T_2=37.485°c$

(b)

Change in enthalpy in kJ;

$\Delta h=mc_p(T_2-T_1)$

$\Delta h=2.5×1531.38(5.485)$

$\Delta u=20.9999kJ$

$\Delta u=Q=21kJ$

(c)

Change in entropy kJ/K;

$\Delta s=mc_pln\frac{T_2}{T_1}$

$\Delta s=2.5×1.53138ln(\frac{310.485}{305})$

$\Delta s=0.0682kJ/K$