Question

Find the values of Cp and ᵞ for a gas whose gas

constant is 0.324 kJ/kg K and Cv = 0 84 kJ/kg K.

If 2.25 kg of this gas with an initial volume of

1.15 m3 undergoes a process during which its

pressure remains constant at 7 bar and its

temperature becomes 280‘C at the end of the

process, calculate: (a) the change in internal

energy, (b) the change in enthalpy, (c) the

transferred heat, and (d) the work done.

Answer 1

Solution;

Given;

$R=0.324kJ/kgK$

$C_v=0.84kJ/kgK$

$m=2.25kg$

$V_1=1.15m^3$

$P_1=P_2=7bar$

$T_2=28°C=301K$

Value of $C_p$ can be calculated as;

$C_p=C_v+R$

$C_p=0.84+0.324=1.164kJ/kgK$

Value for $\gamma$ can be calculated as;

$\gamma=\frac{C_p}{C_v}=\frac{1.164}{0.84}=1.386$

(a)change in internal energy;

$dU=mC_v(T_1-T_2)$

From the relation;

$PV=mRT$

$V_2=\frac{2.25×324×301}{7×10^5}=0.3134m^3$

For ideal gases;

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$T_1=\frac{1.15}{0.3134}×301=1104.5K$

$dU=2.25×840×(1104.5-301)$

$dU=1518.615kJ$

(b)change in enthalpy;

$dH=mC_p(T_1-T_2)$

$dH=2.25×1164×(1104.5-301)$

$dH=2104.366kJ$

(c)Heat transferred;

For isobaric process;

$dH=dQ=2104.366kJ$

(d)Work done;

$W=p\Delta V$

$W=7×10^5(1.15-0.3134)$

$W=585.620kJ$