Answer to Question #273466 in Molecular Physics | Thermodynamics for Neilmar

Question

A certain ideal gas has a constant R = 38.9 ft-lb/lb-°R with k = 1.4. (a) if 3 lb of this gas undergoes a reversible nonflow.constant volume process from P, = 20 psia, 140°F to 740°F find: • the change in internal energy in Btu. . the change in enthalpy in Btu. • the heat in Btu. . the work in Btu. b. If the process in (a) is had been steady flow with AP=O. AK = 0, find: • the work in Btu. . the change in entropy in Btu/R.

Answer 1

Solution;

Given;

$R=38.9ftlb/lb°R$

$k=1.4$

But ;

$778.17ft-lb=1Btu$

Hence;

$R=0.05Btu/lb°R$

(a)

$\frac{c_p}{c_v}=k$

$c_p=1.4c_v$

Also;

$c_p-c_v=R$

$1.4c_v-c_v=0.05$

$c_v=0.125Btu/lb°R$

Mass $=3lb$

$P_1=20psia$

$T_1=140°F=$ $599.67°R$

$T_2=740°F=1199.67°F$

Change in internal energy;

$\Delta u=mc_v(T_2-T_1)$

$\Delta u=3×0.125(1199.67-599.67)$

$\Delta u=225Btu$

Change in enthalpy;

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

$c_p=1.4c_v=0.175$

$\Delta h=3×0.175(1199.67-599.67)$

$\Delta h=315Btu$

The heat,work

The process is constant volume , hence;

$W=0$

From first law of thermodynamics ;

$Q=\Delta u=225Btu$

(b)

The process is steady flow;

$W=v\int dp$

$W=v(p_2-p_1)$

$P_2=\frac{p_2}{T_1}×T_2=\frac{20}{599.67}×1199.67=40.01psia$

$P_2=40.01psia=5761.6lb/ft^2$

$P_1v_1=mRT_1$

$v_1=\frac{3×0.05×599.67}{2880.805}=0.0312ft^3$

$W=0.0312(5761.6-2880.805)$

$W=89.95lb-ft$

$W=0.1156Btu$

Entropy;

$\Delta s=m[c_pln(\frac{T_2}{T_1})-Rln(\frac{p_2}{p_1})]$

$\Delta s=3[0.175ln(\frac{1199.67}{599.67})-0.05ln(\frac{5761.6}{2880.805})]$

$\Delta s=0.2600Btu/°R$