Question

 If 528 kJ of heat are added to 2.25 kg of helium at constant volume process when the initial temperature is 40 deg. C, find:

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;

Q=528kJ

m=2.25kg

$V_1=V_2$

$T_1=40°c=313K$

$C_v=3.12kJ/kgK$

(a)

Final temperature;

$Q=mC_v(T_2-T_1)$

$528=2.25×3.12(T_2-40)$

$T_2=115.21°c$

(b)

change in enthalpy in kJ;

W=0

Hence;

$Q=\Delta h$

$\Delta h=528kJ$

(c)

Change in entropy in kJ/K;

$\Delta s=m[C_vln(\frac{T_2}{T_1}+Rln(\frac{V_2}{V_1})]$

$\Delta s=2.25[3.12ln(\frac{313}{388.21})+0]$

$\Delta s=1.511kJ/K$